solved questions on probability distribution

Turney, S. (2022, November 10). Probability Distribution | Formula, Types, & Examples. Scribbr. Retrieved March 2, 2023, from https://www.scribbr.com/statistics/probability-distributions/

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solved questions on probability distribution

Probability Questions

FACTS AND FORMULAE FOR PROBABILITY QUESTIONS

1. Experiment : An operation which can produce some well-defined outcomes is called an experiment.

2. Random Experiment : An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.

i. Tossing a fair coin.

ii. Rolling an unbiased dice.

iii. Drawing a card from a pack of well-shuffled cards.

3. Details of above experiments:

i. When we throw a coin, then either a Head (H) or a Tail (T) appears.

ii. A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we throw a die, the outcome is the number that appears on its upper face.

iii. A pack of cards has 52 cards.

There are 4 honours of each unit. There are Kings, Queens and Jacks. These are all called face cards.

4. Sample Space: When we perform an experiment, then the set S of all possible outcomes is called the sample space.

1. In tossing a coin, S = {H, T}

2. If two coins are tossed, the S = {HH, HT, TH, TT}.

3. In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}.

Event : Any subset of a sample space is called an event.

5. Probability of Occurrence of an Event :

Let S be the sample and let E be an event.

Then, E ⊆ S

∴ P ( E ) = n ( E ) n ( S )

6. Results on Probability :

i. P(S) = 1 ii. 0 ≤ P ( E ) ≤ 1 iii. P ( ∅ ) = 0

iv. For any events A and B we have :

P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B )

v. If A denotes (not-A), then P ( A ) = 1 - P ( A )

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) = 10 C 2 10 = 10 * 9 2 * 1 =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

= 6 C 2 + 4 C 2 = 6 * 5 2 * 1 + 4 * 3 2 * 1 = 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

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A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

Let A, B, C be the respective events of solving the problem and A , B , C be the respective events of not solving the problem. Then A, B, C are independent event

∴ A , B , C are independent events

Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

P A = 1 2 , P B = 2 3 , P C = 3 4

∴ P( none solves the problem) = P(not A) and (not B) and (not C)

= P A ∩ B ∩ C

= P A P B P C ∵ A , B , C a r e I n d e p e n d e n t

= 1 2 × 2 3 × 3 4

= 1 4

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= 1 - 1 4 = 3/4

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Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

We have n(s) = 52 C 2 52 = 52*51/2*1= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = 52 * 51 2 * 1 = 26 C 2 = 325, n(B)= 26 * 25 2 * 1 = 4*3/2*1= 6 and n(A∩B) = 4 C 2 = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

Two dice are tossed. The probability that the total score is a prime number is:

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

P(E) = n(E)/n(S) = 15/36 = 5/12.

A man and his wife appear in an interview for two vacancies in the same post. The probability of husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected ?

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = 15 C 3 = 15 * 14 * 13 3 * 2 * 1 = 455.

Let E = event of getting all the 3 red balls.

n(E) = 5 C 3 = 5 * 4 2 * 1 = 10.

=> P(E) = n(E)/n(S) = 10/455 = 2/91.

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

P ( E ) = n ( E ) n ( S ) = 10 35 = 2 7

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Solving problems with probability distributions

This post focuses on days 4 and 5 of the “10 Days of Statistics” path in HackerRank, because they are very similar. All tasks can be broken in two parts:

Personally, I prefer the second type of question, as it is more of a real world question: “how do I apply some general knowledge to a particular problem?” The general knowledge is, of course, the probability distribution formula, which you will easily find implemented in some library anyway.

Let’s start then with the creative questions first, and leave the implementation of the distributions to the end. Just assume, for now, that I have appropriately named functions in my ongoing stats.py module.

The essential knowledge for all these questions is the following:

some random event happens with probability dictated by some random distribution

$\Pr[X = x]$

the above function is called the “Probability Mass Function” (PMF)

For some distributions, the PMF is 0 at every point (eg, if the variable can take an infinity of values). Because of this, it’s often more useful to use a Cumulative Distribution Function (CDF).

the Cumulative Distribution Function computes the probability that the random variable takes a value less than or equal to a given value.

$\Pr[X \leq x]$

Where the variable X is discrete, this is the sum of the PMF for the relevant values of X. For real variables, this is the integral of a function known as the Probability Density Function.

Task (day 4) The probability that a machine produces a defective product is 1/3 . What is the probability that the 1st defect is found during the 5th inspection ?

Let us now see how to frame the problem statement as a CDF or PMF calculation. We have a binary event (the product either has a defect or not) that occurs with probability 1/3 (this is all the parameters of the distribution). The problems asks for the probability of the first defective product being detected in the 5th trial . We’d rather break this down. There are two parts here:

$\Pr[X = 5]$

This is all we need to write our code.

Task (day 4) The probability that a machine produces a defective product is 1/3 . What is the probability that the 1st 1 1 defect is found during the first 5 inspections ?

$\Pr[X \leq 5]$

Task (day 5) A random variable, X , follows the Poisson distribution with mean of 2.5 . Find the probability with which the random variable X is equal to 5 .

Task (day 5)

Task The manager of a industrial plant is planning to buy a machine of either type A or type B For each day’s operation:

$C_A = 160 + 40X^2$

Assume that the repairs take a negligible amount of time and the machines are maintained nightly to ensure that they operate like new at the start of each day. Find and print the expected daily cost for each machine.

This is a rather more interesting problem. It feels very school-like, but involves a number of steps, rather than just applying a function directly. In an exam situation, you’d probably be asked which one machine the manager should buy, and let you figure out you’d have to compute the costs for each.

The problem already says we have two Poisson distributions. What it doesn’t say, but is absolutely needed to solve this problem, is that the expected value of a distribution is a linear quantity: the expected value of a sum of random variables is the sum of their expected values.

This should be broken down for clarity:

we know the expected values (the mean) of random variables X and Y

$C_A$

we must return the expected value of each of these new random variables

$E[A + B] = E[A] + E[B]$

In the expression above, we sum two things, but they don’t quite look like A or B . But we can make them do so:

$E[F + A] = E[F] + E[A] = 160 + E[A]$

We now have everything we need to compute the expected costs for each variable.

By the way, the expected cost for machine A is 226.176, and for machine B it is 286.1, so the owner should choose machine A.

Task (day 5) In a certain plant, the time taken to assemble a car is a random variable, X , having a normal distribution with a mean of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in:

Day 5 also introduces the normal distribution, which is probably the most important distribution in statistics. Most of that is because of the Central Limit Theorem, which states that in many cases we can approximate a series of independent observations by a single normal distribution. For this reason, it is often used to estimate the probability distribution of many natural phenomena.

$\Pr[X \leq 19.5]$

As for the second question, notice that if you count the probability of X being less than 22 hours, you also include all the cases in which it is less than 20 . You can break this probability in two sets, as below.

$\Pr[X \leq 22] = \Pr[X \leq 20] + \Pr[20 < X \leq 22]$

Now, it is simple to obtain the solution to the question, by simply rearranging the terms.

The code to these two questions then is:

$\mu = 70$

The cumulative distribution gives us the probability of the variable being at most some value. The probability of a variable being at least a certain value is exactly the complement (remember that for the normal distribution, the probability of being exactly equal to a value is 0). That is,

$\Pr[X \geq x] = 1 - \Pr[X \leq x]$

And this is it for today. I hope the above examples can give you some insight on how to use probability distributions. Some work remains to be done, namely the implementation of the distribution functions themselves. This, I leave for another day.

See you then.

Probability Distribution

What is the Probability Distribution?

Probability distribution could be defined as the table or equations showing respective probabilities of different possible outcomes of a defined event or scenario. In simple words, its calculation shows the possible outcome of an event with the relative possibility of occurrence or non-occurrence as required.

Probability distribution formula, examples of probability distribution formula (with excel template), relevance and uses, recommended articles.

The probability of occurring event can be calculated by using the below formula;

Probability of Event = No of Possibility of Event / No of Total Possibility

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Given below are the examples of the probability distribution equation to understand it better.

Let’s suppose a coin was tossed twice, and we have to show the probability distribution of showing heads.

In the given example, possible outcomes could be (H, H), (H, T), (T, H), and(T, T).

Then possible no. of heads selected will be – 0, or one could calculate 1 or 2, and the probability of such an event by using the following formula:

Calculation of probability of an event can be done as follows,

Probability Distribution Formula Example 1.1

Using the Formula,

Probability of selecting 0 Head = No of Possibility of Event / No of Total Possibility

Probability of an event will be –

Probability of selecting 1 Head = No of Possibility of Event / No of Total Possibility

Probability of selecting 2 heads =No of Possibility of Event / No of Total Possibility

So, the probability distribution for selecting heads could be shown as;

Probability Distribution Formula Example 1.2

Explanation: The event was ‘No. of heads’ in the given example. And the number of heads that can occur is either 0 or 1, or 2, which would be termed as possible outcomes, and the respective possibility could be 0.25, 0.5, 0.25 of the possible outcomes.

In an interview hall, there were 4 people present, consisting of 2 men and 2 women, after being tested by the interviewers. But the concerned company had only 2 vacancies to fill. So, the interviewer decided to select 2 candidates from the people in the hall. So, what will be the probability distribution of ‘selecting at least one woman.’

In the given case, the number of possibilities of selecting candidate could be,

(W1, W2), (W1, M1), (W1, M2), (W2, M1), (W2, M2), (M1, M2)

As per the requirement, let’s denote the event ‘number of women’ as X, then the possible values of X could be;

Calculation of Probability of an event

So, the probability of selecting 0 women = no of the possibility of selecting 1 women / total possibilities

Probability Distribution Formula Example 2.1

Probability of selecting X women = no of the possibility of selecting X women / total possibilities

Probability Distribution Formula Example 2.3

Now, as per the question, the probability of selecting at least 1 woman will be

= Probability of selecting 1 woman + Probability of selecting 2 women

So, the probability distribution for selecting women will be shown as;

Explanation: In this scenario, the management decided to fill up the 2 vacancies through interviews, and during the interview, they chose 4 people. They decide to select randomly for the final selection, and the number of women selected could be either 0 or 1, or 2. The possibility of an event where no women would be selected is, and the possibility of an event where it will select only 1 woman amounted to. In contrast, the possibility of selection of both women is.

So, through probability distribution, the trend of employment, hiring, selection of candidates, and other nature could be summarised and studied.

In the given example, the random variable is the ‘number of damaged tube lights selected.’ So let’s denote the event as ‘X.’

Then, the possible values of X are (0,1,2)

So, one could calculate the probability by using the formula:

Probability of selecting X = no of possibilities of selecting X / total possibilities

Then,Probability of selecting 0 damaged lights = probability of selecting good light in 1 st round X probability of selecting good light in 2 nd round X probability of selecting good light in 3 rd round.

(multiplied by 3 because the damaged light can be selected in 3 ways, i.e., either in 1 st round or 2 nd or 3 rd round)

(multiplied by 3 because the good light can be selected in 3 ways, i.e., either in 1 st round or 2 nd or 3 rd round)

P (2) = (8/10*2/9*1/8)*3

So the probability of selecting at least 1 Damaged lights = Probability of selecting 1 Damage + Probability of selecting 2 Damage

= P (1) + P (2)

So, the probability distribution for selecting damage lights could be shown as;

Explanation: The Operation Manager of the business organization wanted to evaluate the effectiveness of the process by randomly selecting goods and evaluating the chances of producing damaged goods.

Through this example, we can see that the industry can also use Probability distribution to evaluate the effectiveness of its processes and the ongoing trends.

One may use a probability distribution for recording the possibility of the occurrence or non-occurrence of a certain event. From a business point of view, one can also use it for predicting or estimating the possible future returns or profitability of the business. In modern-day business, the probability distribution calculation is for sales forecasting, risk evaluation, finding and evaluating the obsolete part of any business or process, etc.

This article is a guide to Probability Distribution and its meaning. Here, we discuss the formula to calculate probability distribution, practical examples, and a downloadable Excel template. You can learn more from the following articles: –

Probability Questions And Answers

The branch of mathematics that deals with the likelihood of occurrence of an event is termed probability. The range of probability is between 0 and 1. 0 indicates the impracticality of the happening of an event and 1 indicates the certainty of happening of an event. The more the probability the more likely the event to occur. In tossing a coin, the end results are head or tail. Both the results have an equal possibility of occurring. The probability of heads or tails is 0.5.

Probability Formula

The probability of any event E is given by the ratio of the count of the favourable outcomes of the event to the total number of possible outcomes of a random experiment.

P (an event) = count of favourable outcomes / total count of outcomes

Solved Probability Examples

Example 1: The tickets are marked from number 1 to 20. One ticket is chosen at random. Find the probability that the ticket selected has a digit that is a multiple of number 3 or number 5.

Answer:

The sample space of the above problem is = S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

Multiples of 3 are {3, 6, 9, 12, 15, 18}.

Multiples of 5 are {5, 10, 15, 20}.

P (the ticket selected has a digit that is a multiple of number 3 or number 5) = (multiples of 3 or multiples of 5) / total number of possible outcomes

= {3, 6 , 9, 12, 15, 18, 5, 10, 20} / 20

Example 2: There are 2 red coloured, 3 green coloured and 2 blue coloured balls in a bag. Two balls need to be drawn at random. Find the probability that none of the balls drawn is blue in colour.

Total number of balls = 2 red + 3 green + 2 blue = 7 balls

S is the sample space.

Two balls need to be drawn at random.

n (S) = number of methods of drawing 2 balls out of 7 balls

Let E be the event of drawing 2 balls such that none of the balls drawn is blue in colour.

Number of blue balls = 2

Number of other coloured balls = 5

n (E) ={ }^{5} \mathrm{C}_{2} \\ =\frac{(5 \times 4)}{(2 \times 1)} \\ =10

P (E) = n (E) / n (S)

Example 3: There are 300 students in a school. Out of them, 95 students play cricket, 120 students play football, 80 students play volleyball and 5 students don’t play any games. A student is chosen randomly, find the probability that the chosen student

a] plays volleyball

b] either cricket or volleyball sport

c] neither football nor volleyball

Total number of students = 300

95 students play cricket.

120 students play football.

80 students play volleyball.

5 students don’t play any games.

P (chosen student plays volleyball) = number of students playing volleyball / total number of students

P (chosen student plays either cricket or volleyball) = (sum of the students who play cricket and volleyball) / total number of students

= (95 + 80) / 300

= 175 / 300

P (chosen student plays neither football nor volleyball) = (total number of students – number of students who play football – number of students who play volleyball) / total number of students

= (300 – 120 – 80) / 300

= 100 / 300

Example 4: Two fair dice are rolled. What is the probability of the following events?

a] probability that the sum is 1.

b] probability that the sum is 4.

c] probability that the sum is less than the number 13.

The sample space when two fair dice are rolled is given below.

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

There are no outcomes from the sample that gives the sum as 1.

P (the sum is 1 when two fair dice are rolled) = 0 / 36 = 0

The outcome that gives the sum as 4 = {(1, 3), (2, 2), (3, 1)} = 3 outcomes

P (the sum is 4 when two fair dice are rolled) = 3 / 36

All the outcomes in the sample space give the sum that is less than the number 13.

P (the sum is less than the number 13 when two fair dice are rolled) = 36 / 36

Example 5: There are 90 discs that are numbered from 1 to number 90 in a box. 1 disc is selected at random from the box given. Find the probability that

a] 2-digit numbers are obtained

b] perfect squares are obtained

c] multiples of number 5 are obtained

d] numbers that are divisible by 3 and 5 are obtained

a] a 2-digit number is obtained

Total number of possible outcomes = 90 (given)

Let E be the event of getting a 2-digit number.

Number of 2-digit numbers from 1 to 90 = 90 – 9 = 81 (the single-digit numbers are from 1 to 9).

P (a 2-digit number is obtained) = Number of 2-digit numbers from 1 to 90 / total number of outcomes

= 81 / 90

b] a perfect square is obtained

Let F be the event of obtaining perfect squares.

The number of perfect squares from 1 to 90 = {1, 4, 9, 16, 25, 36, 49, 64 and 81} = 9

P (F) = count of perfect squares from 1 to 90 / total number of possible outcomes

Let G be the event of obtaining multiples of number 5.

The number of multiples of number 5 from 1 to 90 is 18.

P (G) = P (multiples of number 5 are obtained) = number of multiples of number 5 from 1 to 90 / total number of possible outcomes

Let H be the event of obtaining the numbers that are divisible by 3 and 5.

The count of the numbers divisible by 3 and 5 is 6.

P (H) = P (numbers that are divisible by 3 and 5 are obtained) = 6 / 90

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Probability | Theory, solved examples and practice questions

When MS and MBA applicants ask us – ‘ What are my chances of getting into Harvard? ‘ or ‘ What’s my probability of getting scholarships from Oxford? ‘ we get tongue-tied. There are so many variables at play, it’s difficult to give an accurate answer.

But when you get probability questions in your GRE and GMAT exam syllabus , you don’t have to get flummoxed. Understanding the basic rules and formulas of probability will help you score high in the entrance exams.

Meaning and definition of Probability

As the Oxford dictionary states it, Probability means ‘The extent to which something is probable; the likelihood of something happening or being the case’.

In mathematics too, probability indicates the same – the likelihood of the occurrence of an event.

Examples of events can be :

Either an event will occur for sure, or not occur at all. Or there are possibilities to different degrees the event may occur.

An event that occurs for sure is called a Certain event and its probability is 1.

An event that doesn’t occur at all is called an impossible event and its probability is 0.

This means that all other possibilities of an event occurrence lie between 0 and 1.

This is depicted as follows:

0 <= P(A) <= 1

where A is an event and P(A) is the probability of the occurrence of the event.

This also means that a probability value can never be negative.

Every event will have a set of possible outcomes. It is called the ‘sample space’.

Consider the example of tossing a coin.

When a coin is tossed, the possible outcomes are Head and Tail. So, the sample space is represented as {H, T}.

The probability of head each time you toss the coin is 1/2. So is the probability of tail.

Basic formula of probability

As you might know from the list of GMAT maths formulas , the Probability of the occurrence of an event A is defined as:

P(A) = (No. of ways A can occur)/(Total no. of possible outcomes)

Another example is the rolling of dice. When a single die is rolled, the sample space is {1,2,3,4,5,6}.

What is the probability of rolling a 5 when a die is rolled?

No. of ways it can occur = 1

Total no. of possible outcomes = 6

So the probability of rolling a particular number when a die is rolled = 1/6.

Compound probability

Compound probability is when the problem statement asks for the likelihood of the occurrence of more than one outcome.

Formula for compound probability

P(A or B) = P(A) + P(B) – P(A and B)

where A and B are any two events.

P(A or B) is the probability of the occurrence of atleast one of the events.

P(A and B) is the probability of the occurrence of both A and B at the same time.

Mutually exclusive events:

Mutually exclusive events are those where the occurrence of one indicates the non-occurrence of the other

When two events cannot occur at the same time, they are considered mutually exclusive.

Note: For a mutually exclusive event, P(A and B) = 0.

Example 1: What is the probability of getting a 2 or a 5 when a die is rolled?

Taking the individual probabilities of each number, getting a 2 is 1/6 and so is getting a 5.

Applying the formula of compound probability,

Probability of getting a 2 or a 5,

P(2 or 5) = P(2) + P(5) – P(2 and 5)

==> 1/6 + 1/6 – 0

==> 2/6 = 1/3.

Example 2: Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

We need to find out P(B or 6)

Probability of selecting a black card = 26/52

Probability of selecting a 6 = 4/52

Probability of selecting both a black card and a 6 = 2/52

P(B or 6) = P(B) + P(6) – P(B and 6)

= 26/52 + 4/52 – 2/52

Independent and Dependent Events

Independent event.

When multiple events occur, if the outcome of one event DOES NOT affect the outcome of the other events, they are called independent events.

Say, a die is rolled twice. The outcome of the first roll doesn’t affect the second outcome. These two are independent events.

Example 1: Say, a coin is tossed twice. What is the probability of getting two consecutive tails ?

Probability of getting a tail in one toss = 1/2

The coin is tossed twice. So 1/2 * 1/2 = 1/4 is the answer.

Here’s the verification of the above answer with the help of sample space.

When a coin is tossed twice, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

Our desired event is (T,T) whose occurrence is only once out of four possible outcomes and hence, our answer is 1/4.

Example 2: Consider another example where a pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 blue pens and 1 black pen?

Here, total number of pens = 9

Probability of drawing 1 blue pen = 4/9 Probability of drawing another blue pen = 4/9 Probability of drawing 1 black pen = 3/9 Probability of drawing 2 blue pens and 1 black pen = 4/9 * 4/9 * 3/9 = 48/729 = 16/243

Dependent Events

When two events occur, if the outcome of one event affects the outcome of the other, they are called dependent events.

Consider the aforementioned example of drawing a pen from a pack, with a slight difference.

Example 1: A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from the pack, NOT replaced and then another pen is drawn. What is the probability of drawing 2 blue pens and 1 black pen?

Probability of drawing 1 blue pen = 4/9 Probability of drawing another blue pen = 3/8 Probability of drawing 1 black pen = 3/7 Probability of drawing 2 blue pens and 1 black pen = 4/9 * 3/8 * 3/7 = 1/14

Let’s consider another example:

Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement.

Probability of drawing a king = 4/52 = 1/13

After drawing one card, the number of cards are 51.

Probability of drawing a queen = 4/51.

Now, the probability of drawing a king and queen consecutively is 1/13 * 4/51 = 4/663

Conditional probability

Conditional probability is calculating the probability of an event given that another event has already occured .

The formula for conditional probability P(A|B), read as P(A given B) is

P(A|B) = P (A and B) / P(B)

Consider the following example:

Example: In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?

P(M and S) = 0.40

P(M) = 0.60

P(S|M) = P(M and S)/P(S) = 0.40/0.60 = 2/3 = 0.67

Complement of an event

A complement of an event A can be stated as that which does NOT contain the occurrence of A.

A complement of an event is denoted as P(A c ) or P(A’).

P(A c ) = 1 – P(A)

or it can be stated, P(A)+P(A c ) = 1

For example,

if A is the event of getting a head in coin toss, A c is not getting a head i.e., getting a tail.

if A is the event of getting an even number in a die roll, A c is the event of NOT getting an even number i.e., getting an odd number.

if A is the event of randomly choosing a number in the range of -3 to 3, A c is the event of choosing every number that is NOT negative i.e., 0,1,2 & 3 (0 is neither positive or negative).

Example: A single coin is tossed 5 times. What is the probability of getting at least one head?

Consider solving this using complement.

Probability of getting no head = P(all tails) = 1/32

P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.

Sample Probability questions with solutions

Probability example 1.

What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled.

Let the event of the occurrence of a number that is odd be ‘A’ and the event of the occurrence of a number that is less than 5 be ‘B’. We need to find P(A or B).

P(A) = 3/6 (odd numbers = 1,3 and 5)

P(B) = 4/6 (numbers less than 5 = 1,2,3 and 4)

P(A and B) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)

Now, P(A or B) = P(A) + P(B) – P(A or B)

= 3/6 + 4/6 – 2/6

P(A or B) = 5/6.

Probability Example 2

A box contains 4 chocobars and 4 ice creams. Tom eats 3 of them one after another. What is the probability of sequentially choosing 2 chocobars and 1 icecream?

Probability of choosing 1 chocobar = 4/8 = 1/2

After taking out 1 chocobar, the total number is 7.

Probability of choosing 2nd chocobar = 3/7

Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3

So the final probability of choosing 2 chocobars and 1 icecream = 1/2 * 3/7 * 2/3 = 1/7

Probability Example 3

When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8.

Let the event of getting a greater number on the first die be G.

There are 5 ways to get a sum of 8 when two dice are rolled = {(2,6),(3,5),(4,4), (5,3),(6,2)}.

And there are two ways where the number on the first die is greater than the one on the second given that the sum should equal 8, G = {(5,3), (6,2)}.

Therefore, P(Sum equals 8) = 5/36 and P(G) = 2/36.

Now, P(G|sum equals 8) = P(G and sum equals 8)/P(sum equals 8)

= (2/36)/(5/36)

Probability Quiz: Sample probability questions for practice

Problem 1: Click here

A bag contains blue and red balls. Two balls are drawn randomly without replacement. The probability of selecting a blue and then a red ball is 0.2. The probability of selecting a blue ball in the first draw is 0.5. What is the probability of drawing a red ball, given that the first ball drawn was blue? a) 0.4 b) 0.2 c) 0.1 d) 0.5

Answer 1: Click here

Problem 2: Click here

A die is rolled thrice. What is the probability that the sum of the rolls is atleast 5. a) 1/216 b) 1/6 c) 3/216 d) 212/216

Answer 2: Click here

Learn how to solve: – Simple and compound interest problems – Speed, distance and time problems – Ratio and proportion – List of Maths Formulas

43 thoughts on “Probability | Theory, solved examples and practice questions”

a number of people gave a hat check girl one hat. suppose all the tickets got misplaced, so all the hat were given back randomly. a) if its 2 people determine the probability at least one person got their hat returned. b) if its 3 people determine the probability at least one person got their hat returned. c) if its 4 people determine the probability at least one person got their hat returned. d) if its 5 people determine the probability at least one person got their hat returned.

Hi I’m Algia and I need help in solving this problem, can you help me please.

a) 1-(0.5)=0.5 b) 1-(0.667*0.5)=0.667 c) 1-(0.75*0.667*0.5)=0.75 d) 1-(0.8*0.75*0.667*0.5)=0.8

Really nice sequence.

P(h’)=1-P(h) etc. At least one hat is correctly returned is compliment that no hat is returned correctly.

The math here is totally wrong

Feel free to enlighten!

you make the assumption the events are independent, they are not, if you there are only 2 people, it is impossible for just 1 person to be given the right hat, it must be both people.

You make the assumption of independence, which does not hold in this question. instead we have to use the inclusion-exclusion principle.

For a) the probability that at least 1 person gets the got their hat returned. A is the event the 1st person gets their hat and B is the event the 2nd person gets their hat. We want to find P(A U B) = ?

P(A U B) = P(A) + P(B) – P(A n B)

P(A) = 1/2 P(B) = 1/2 P(A n B) = 1/4

P(A U B) = 1/2 + 1/2 – 1/4 = 3/4

Think it in this way. Representation here: Got Hat – G, Didn’t Get Hat – D. If there are 2 people, the total possible outcomes would be = 2^2 = 4 with them being {(G,G),(D,D),(G,D),(D,G)}. Now, probability that at least one person got hat returned = 1 – P(no one gets hat) = 1 – 1/4 = 3/4.

and so on…

General Formula derived for ‘n’ people case : P(At least one person gets hat returned) = 1 – P(no one gets hat) = (1 – 1/2^n)

Hi Last question must be 212/216 right ?

Tell me the way u did that sol. Plzz…

yes its must be 212/216

I think it should be 212/216 Bcz we have 4 number of event to getting a number of sum less than 5 {(1,1,1),(1,1,2),(1,2,1),(2,1,1)} It means p(e) = 4/216 Nd getting a number of sum at least 5 Is 1-4/216=212/216

you are telling a wrong way so stop telling me your bloody wrong answer it’s a correct answer it’s should not be 212/216..

@Emmanuel, I don’t think you can isolate the two variables like that since, as you mentioned, these are dependent events. The total number of ways of returning the hat is n! and the total number of ways of returning the hats such that no one gets their own one is (n-1)!. Therefore, the probability that at least one person gets their own hat is P(X>0) = 1-P(X=0) = 1 – (n-1)!/n! = 1 – 1/n.

A woman bought 5basket of tomatos each costing 1250naira,in her discovery she observe that 90% of the tomatos where damage resulting to a loss of 510naira.(a)what is the probability of obtaining an average of 50 if the cost per bag is 50 above the cost?(b)what will be the actual price for selling the tomato at cost plus(+) 25%?

Simple way: Q: A die is rolled thrice. What is the probability that the sum of the roll is at least 5?

A: At least 5 is equal or greater 5 ===> which is P(x> or =5)= 1-P(x<5) Then: P(x<5)= 4/216. . .taken thrice rolls =6*6*6=216

Therefore: 1-P(x 53/54

This is incorrect. Correct answer is: P(x<5) = 3 Therefore Answer = 1 – 3/216 = 213/216.

the personal director of a company wishes to select applicant for advanced training without regard to sex. let “W” denotes women and “M” denotes men and the pattern of arrival be M WWW MMM WW M WWW MMMM W M W MM WWW MM W MMMM WW M WW MMMM WW M WWWW MM WW M W WW. will you conclude that the applicants have arrived in a random fashion?

pavement.before any 250 m length of a pavement is accepted by the state highway department,the thickness of a30 m s mointored by an altrasonic to verify compliance to specification .each section is rejected if a measurment thickness less than 10cm;otherwise the all section is accepted .from past experment ,the stat highway engineer know the 85%of all section constructed by the contructor comply with specification . however the relability of altrusonic thickness testing is only 75 ,so that there is a 25 percent chane of errorneous concolusion based on the determenation of thickness with ultrasonic . what is the probablity that a poorly constructed section is accepted on the base of ultrasonic test?

The chance or probability of getting accepted is 0.85; the chance of getting accepted even when bad is 0.25. So therefore the chance of being bad and getting selected can be solved using the conditional probability theorem given by:

P(A/B)= P(AnB)/P(B). Going by this the answer is: 0.25 x 0.85= 0.2125

solution the possible out come of rolling die is =6 here in this case since it is rolled 3 our sample space is 6×6×6=216 we have asked to solve the probability of sum which will be atleast 5 this means 5 and more is possible. so that we have to search the possibilities of less than five to easy our work this will be like[111][112][121] = 3 out comes onlywso p(s`)=3/216 when p(s`) is probability of sum less than five or probability of sum greater than equal to five. since the sum of p(s) and p(s`)=1 p(s)=1-p(s~) 1-3/216=213/216

what about 2,1,1?

Two cards are drawn at random from an ordinary deck of 52 card. Find the probability P that (a) Both are spade (b) One is a spade and one is heart

Ans: (a) Probability of getting spade 1st time is 13/52 and Probability of getting spade 2nd time is is 12/51 Total Probability is 13*12/(52*51) = 156/2652 (b) Probability of getting spade is 13/52 and Probability of getting Heart is 12/51 Total probability is 13*13/(52*51) = 169/2652

Question ‘b’ says that one is a spade and one is a heart. Therefore the possibilities are ‘heart-spade’ and ‘spade-heart’. The answer should be: ((13*13/52*51) + (13*13/52*51)) = 169/2652 + 169/2652 =338/2652 = 13/102

copying the solution offerred by @ diriba

The above solution is good but a little faulty because it considered only the possibility of obtaining a ‘1’ on the first die, it omitted the possibility of getting a ‘2’ on the first die i.e (using the same notation) [211], this is the fourth possible outcome. Hence P(s)= 1- P(s’) = 1-4/216 =212/216 =53/54

Lets assume probability of picking a red ball is X. The probability of selecting a blue ball and then a red ball, P(B)*P(R)=.2 .5*X=.2 x=.5/.2 x=.4

P(R|B)=P(R and B)/P(B) =0.2/0.5=0.4

The probability of snow tomorrow is 0.6. And the probability that it will bi colder is 0.7. The probability that it will not snow and not bi colder is 0.1 .What is probability that it will not snow if it is colder tomorrow? please solve it … and tell me answer.. thanks …

A= event for it will snow tomorrow B= event for it will be cold a= event for it will not snow b= event for it will not be cold P(A) = 0.6; P(B)= 0.7; P(a n b)=0.1 P(A)+P(B)-P(AnB)+P(a n b)=1 Inserting values, you’ll have P(AnB)=0.4. P(Bna)= P(B)-P(BnA)

This will give you P(Bna)= 0.3.

By conditional theorem,

P(a/B)=P(anB)/P(B)

This will give you 3/7 as the answer.

pl first draw a tree diagram: P(S)=0.6, P(N.S.)=0.4 P(Cold)=0.7, P(Not Cold)=0.3 P(No S + Not Cold)= 0.4*0.3=0.12=0.1 P(No S + Cold)=0.4*0.7=0.28

in a class 10 boys and 5 girls .three students are selected random one after the other.find the probability that

1)first two are boys and third is girl 2)first and third is of same gender and third is of opposite gender

please help me in solving this

A) 10/15*9/14*5/13 B) 1st case: 1st & 2nd are boys & 3rd is girl 10/15*9/14*5/13 2nd case: 1st & 2nd are girls & 3rd is boy 5/15*4/14*10/13

n(b)=10 n(g)=5 n(t)=15 (a) p(first two are boys and third is a girl) p(B,B,G) 10/15*9/14*5/13= 5/1*3/7*1/13=15/91 (b) p(first and third is of the same gender and second is opposite) P(B,G,B) or p(G,B,G) (10/15*5/14*9/13)+(5/15*10/14*4/13)= (450/2730)+(200/2730)= 450+200/2730= 650/2730= 5/21

There are three boxes, one of which contains a prize. A contestant is given two chances, such that if he chooses the wrong box in the first round, that box is removed from the selection and he then chooses between the two remaining boxes. 1. What is the probability that the contestant wins? 2. Does the contestant’s probability of winning increases on the second round?

It’s a Monty Hall problem. You can google it.

As for your question, As the first box chosen if found empty is removed and you HAVE/Switch to pick from other two, the P(W) = 2/3.

Above answer can be explained as Prob. of winning on first box + Prob. of choosing wrong * Prob. of Choosing right between the two => 1/3+2/3*1/2 => 2/3

The answer to the second: Yes probability increases as its a 50% chance to win as 1 wrong box is eliminated.

1) 10C2*5C1/15C3? 2) (10C1*5C1*9C1/15C3) + (5C1*10C1*4C1/15C3)?

Plz solve it

XYZ company wants to start a food outlet in pakistan. There is a 40% and 60% chance of stating in hyderabad and karachi respectively. If he start the outlet in hyderabad there is 30% chance that it will be in saddar and 70% chance that it will be in defence area. If they start the outlet in karachi there is 50% chance that it will be in defence, 30% in clifton and 20% in pechs. Determine probability of starting the outlet in: (a) saddar (b) defence area of any city (c) clifton given that the outlet is started in karachi

a) P(H,S) = 40% x 30% =0.4 x 0.3 = 0.12 = 12% b) P(H, D) + P(K, D) = 40% x 70% + 60% x 50% = 0.4 x 0.7 + 0.6 x 0.5 = 0.28 + 0.3 = 0.58 = 58% c) P(C|K) = 30%

In maternity clinic the probability of new born was females is 55%=0.55

So,the probabilitt of the next three deliveries are females is 0.55×0.55×0.55=0.166 or 16.6%

Nah, that’s not correct. Births are independent events. It doesn’t matter what the gender of previous births were. The odds of the next birth being female is still 50%. The probability of three females in a row is simply (0.5)^3 = 1/8 = 0.125

a) 15*(0.2)^4*(0.8)^2 = 0.01536 b) 0.01536 + 6*(0.2)^5*(0.8)^1 + 1*(0.2)^6*(0.8)^0 = 0.01696 c) (0.8)^6 = 0.262144

hopefully you haven’t been waiting over a year for a response 🙂 Here you go though…

A: (2/3)^ 3 = 8/27 B: 1 * 2/3 * 1/3 = 2/9 C: 1 * 1/3 * 1/3 = 1/9

Free Mathematics Tutorials

Poisson probability distribution examples and questions.

Poisson probability distribution is used in situations where events occur randomly and independently a number of times on average during an interval of time or space. The random variable $ X $ associated with a Poisson process is discrete and therefore the Poisson distribution is discrete.

Poisson Process Examples and Formula

Example 1 These are examples of events that may be described as Poisson processes:

Compare Binomial and Poisson Distributions

Example 6 The number of defective items returned each day, over a period of 100 days, to a shop is shown below.

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Probability distribution mcqs | probability distribution multiple choice questions and answers.

Free download in PDF Probability Distribution Multiple Choice Questions and Answers for competitive exams. These short objective type questions with answers are very important for Board exams as well as competitive exams. These short solved questions or quizzes are provided by Gkseries.

View Answer

Answer: Only if random variables exhibit statistical independency

Answer: P (G) = 1 – P (H)

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Answer: Continuous random variable

Answer: P(x) = – 0.5

Answer: the area under the curve between points a and b represents the probability that X = a

Answer: 0.50

Answer: The mean of the Poisson distribution (with parameter μ) equals the mean of the Exponential distribution (with parameter λ) only when μ = λ = 1

Answer: flatter and wider

Answer: Poisson Distribution

Answer: Does not contain any common sample point

Answer: Probability Distribution

Answer: Discrete random variable

Answer: the mean is always zero

Answer: both approach zero as x approaches infinity

Answer: Exponential

Answer: Binomial distribution

Answer: all of the above statements are true

Answer: The number of customers arriving at a petrol station

Answer: the density function is constant for all values that X can assume

Answer: the mean of the exponential distribution is the inverse of the mean of the Poisson

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Binomial Distribution Questions

Binomial distribution questions for Class 12 with solutions are provided here for practice. Before understanding the concept of the binomial distribution, let us understand some facts about binomial experiments. There are some sorts of experiments which have only two possible outcomes, either a “ success ” or a “ failure ” – these types of random experiments are called binomial experiments or “ Bernoulli trials ”. For example, the experiment of tossing a coin and getting a head.

Thus, in a probability distribution , binomial distribution denotes the success of a random variable X in an n trials binomial experiment. Following are the conditions to find binomial distribution:

Bernoulli’s Theorem for Binomial Distribution

Let there be ‘n’ binomial experiment trials and let the random variable X denote the success of these trials. If p is the probability of success and 1 – p = q is the probability of failure in each trial, then,

As P(X) is the term of the binomial expansion of (p + q) n , it is called the binomial distribution.

Learn more about binomial distribution .

Binomial Distribution Questions with Solutions

Let us practice some important questions on binomial distribution in probability.

Question 1:

Find the binomial distribution of getting a six in three tosses of an unbiased dice.

Let X be the random variable of getting six. Then X can be 0, 1, 2, 3.

Here, n = 3

p = Probability of getting a six in a toss = ⅙

q = Probability of not getting a six in a toss = 1 – ⅙ = ⅚

P(X = 0) = n C r p r q (n – r) = 3 C 0 (⅙) 0 (⅚) 3 – 0

= 1 × 1 × 125/216 = 125/216

P(X = 1) = n C r p r q (n – r) = 3 C 1 (⅙) 1 (⅚) 3 – 1

= 3 × ⅙ × 25/36 = 25/72

P(X = 2) = n C r p r q (n – r) = 3 C 2 (⅙) 2 (⅚) 3 – 2

= 3 × 1/36 × ⅚ = 5/72

P(X = 3) = n C r p r q (n – r) = 3 C 3 (⅙) 3 (⅚) 3 – 3

= 1 × 1/216 × 1 = 1/216

The required binomial distribution of X is:

Question 2:

Find the probability distribution of the number of doublets in four throws of a pair of dice.

There are 36 total possible outcomes for a throw of dice, for which the following outcomes are the success of the experiment: {(1,1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

p = probability of getting doublets = 6/36 = ⅙

q = probability of getting not getting doublets = 1 – ⅙ = ⅚

X: numbers of doublets, then X = 0, 1, 2, 3, and 4.

P(X = 0) = n C r p r q (n – r) = 4 C 0 (⅙) 0 (⅚) 4 – 0

= 1 × 1 × 625/1296

P(X = 1) = n C r p r q (n – r) = 4 C 1 (⅙) 1 (⅚) 4 – 1

= 4 × ⅙ × 125/216 = 125/324

P(X = 2) = n C r p r q (n – r) = 4 C 2 (⅙) 2 (⅚) 4 – 2

= 6 × 1/36 × 25/36

P(X = 3) = n C r p r q (n – r) = 4 C 3 (⅙) 3 (⅚) 4 – 3

= 4 × 1/216 × ⅚ = 20/1296

P(X = 4) = n C r p r q (n – r) = 4 C 4 (⅙) 4 (⅚) 4 – 4

= 1 × 1/1296 × 1 = 1/1296.

∴ The required probability distribution is:

Question 3:

Find the probability of getting at least 5 times head-on tossing an unbiased coin for 6 times by using the binomial distribution.

p = P(getting an head in a single toss) = ½

q = P(not getting an head in a single toss) = ½

X = successfully getting a head

P(X ≥ 5) = P(getting at least 5 heads) = P(X = 5) + P(X = 6)

= 6 C 5 (½) 5 (½) (6 – 5) + 6 C 6 (½) 6 (½) 6 – 6

= 6 × (½) 6 + 1 × (½) 6 = 7/24.

Hence, the probability of getting at least 5 heads is 7/24.

Question 4:

There are four fused bulbs in a lot of 10 good bulbs. If three bulbs are drawn at random with replacement, find the probability of distribution of the number of fused bulbs drawn.

This is a problem of binomial distribution as the event of drawing a fused bulb is independent.

p = P(drawing a fused bulb) = 4/(10 + 4) = 2/7

q = P(drawing a bulb which is not fused) = 1 – 2/7 = 5/7

X = event of drawing a fused bulb

X can take up the values 0, 1, 2, 3

P(X = 0) = P(getting zero fused bulbs in all draws)

= n C r p r q (n – r)

= 3 C 0 (2/7) 0 (5/7) (3 – 0)

= 1 × 1 × (125/343) = 125/343

P(X = 1) = P (getting one time fused bulb)

= 3 C 1 (2/7) 1 (5/7) (3 – 1)

= 3 × (2/7) × (25/49) = 150/343

P(X = 2) = P(getting two times fused bulbs)

= 3 C 2 (2/7) 2 (5/7) (3 – 2)

= 3 × 4/49 × (5/7) = 60/343

P(X = 3) = (P(getting three times fused bulb)

= 3 C 3 (2/7) 3 (5/7) (3 – 3)

= 1 × 8/343 × 1 = 8/343

The required probability distribution:

Question 5:

On average, every one out of 10 telephones is found busy. Six telephone numbers are selected at random. Find the probability that four of them will be busy.

Let X: event of getting a busy phone number

p = P(probability of getting a phone number busy) = 1/10

q = P(probability of not getting a phone number busy) = 9/10

The required probability = P(X = 4) = 6 C 4 p 4 q (6 – 4)

= 15 × (1/10) 4 × (9/10) 2

= 15 × 81/10 6

= 0.001215.

Question 6:

An unbiased dice is thrown until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw.

Since each throw is independent of the previous throws, we can apply the binomial distribution formula to find the probability.

p = P(getting a six in a throw) = ⅙

q = P(not getting a six in a throw) = 1 – ⅙ = ⅚

According to the question, two sixes are already obtained in the previous throws.

∴ Required probability = P(getting exactly two sixes in five throws) × P(getting a six in the sixth throw)

= 5 C 2 p 2 q 3 × 1 C 1 p 1 q 1 – 1

= 10 × (⅙) 2 × (⅚) 3 × 1 × (⅙)

= 10 × (⅙) 3 × (⅚) 3

= 625/23328.

Question 7:

The probability of a boy guessing a correct answer is ¼. How many questions must he answer so that the probability of guessing the correct answer at least once is greater than ⅔?

p = P(guessing a correct answer) = ¼

q = P(not guessing a correct answer) = ¾

Let him answers n number of questions, then

P(X ≥ 1) = P(guessing at least one correct answer out of n questions) = 1 – P(no success) = 1 – q n

Given, 1 – q n > ⅔ ⇒ 1 – (¾) n > ⅔

⇒ (¾) n < ⅓

Now, let us check the above inequality for different values of n = 1, 2, 3, 4, …

Thus, he must answer at least 4 questions.

Question 8:

When a biased coin is tossed, the probability of getting a head 3 times more than the probability of getting a tail. Find the probability distribution for getting a tail, if the coin is tossed twice.

Let the probability of getting a tail be p, then the probability of getting a head will be 3p

Now, p + 3p = 1 ⇒ p = ¼

q = P(not getting a tail) = 1 – ¼ = ¾

X = event of getting a tail in a toss

Then, possible values of x will be 0, 1, 2

P(X = 0) = 2 C 0 p 0 q 2 – 0

= 1 × 1 × (¾) 2

P(X = 1) = 2 C 1 p 1 q 2 – 1

= 2 × (¼) × (¾)

P(X = 2) = 2 C 2 p 2 q 2 – 2

= 1 × (¼) 2 × 1

The probability distribution for getting the tail is:

Also try: Binomial Distribution Calculator

Question 9:

A bag contains 5 green balls and 3 red balls. If two balls are drawn from the bag randomly with replacement, find the probability distribution of the number of green balls drawn.

Let p = P(getting a green ball) = 5/(5 + 3) = 5/8

q = P(not getting a green ball) = 1 – 5/8 = 3/8

X = event of drawing the green ball, then the value of X could be 0, 1, 2

P(X = 0) = Probability of getting no green ball = 2 C 0 p 0 q 2 – 0 = 1 × 1 × (3/8) 2 = 9/64

P(X = 1) = Probability of getting one green ball = 2 C 1 p 1 q 2 – 1 = 2 × (⅝) × (⅜) = 15/32

P(X = 2) = Probability of getting 2 green balls = 2 C 2 p 2 q 2 – 2 = 1 × (⅝) 2 × (⅜) 0 = 25/64

The required probability distribution is:

Question 10:

Find the probability distribution of getting the number of fours in three throws of a dice. Also, find the mean and variance of the distribution.

Let, p = P(getting a four in a throw of dice) = ⅙

q = P(not getting a four in a throw of dice) = ⅚

X: number of four obtained, then the value of X could be 0, 1, 2, 3.

P(X = 0) = 3 C 0 p 0 q 3 – 0 = 1 × (⅚) 3 = 125/216

P(X = 1) = 3 C 1 p 1 q 3 – 1 = 3 × (⅙) × (⅚) 2 = 75/216

P(X = 2) = 3 C 2 p 2 q 3 – 2 = 1 × (⅙) 2 × (⅚) 3 – 2 = 15/216

P(X = 3) = 3 C 3 p 3 q 3 – 3 = 1 × (⅙) 3 × (⅚) 3 – 3 = 1/216

The required probability distribution

Mean = np = 3 × ⅙ = 1.2

Variance = npq = 3 × ⅙ × ⅚ = 5/12.

Practice Problems on Binomial Distribution

1. Find the binomial distribution of getting an even number if an unbiased dice is thrown thrice.

2. How many times must a man toss an unbiased coin to get at least one head is more than 90%?

3. Three cards are drawn with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces drawn. Also, find the mean and variance of the distribution.

4. The probability of a trainee archer hitting the target is ¼. If he takes 7 shots, what is the probability of his hitting the target at least twice?

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Tricks To Solve Probability Questions

Last Updated : 02 Dec, 2022

The application or uses of probability can be seen in quantitative aptitude as well as in daily life. It is needful to learn the basic concept of probability. We will cover the basics as well as the hard level problems for all levels of students for all competitive exams especially SBI PO, SBI CLERK, IBPS PO, IBPS CLERK, RRB PO, NICL AO, LIC AAO, SNAP, MAT, SSC CGL etc.

Definition:

Probability means the possibility or chances of an event occurring or happening. For example, when a coin is tossed, then we will get ahead or tail. It is a state of probability.

In an event, the happening probability is equal to the ratio of favourable outcomes to the total number of possible outcomes. It represents as,

Number of favourable outcomes = ____________________________________ Total number of possible outcomes

Sample Space:-

It is a set of all possible outcomes of an experiment. It is denoted by S. For example, the Sample space of a die, S = [ 1, 2, 3 , 4, 5, 6] The Sample space of a coin, S= [ Head, Tail]

Types of questions asked in the competitive exam:

1) Based on Coins

2) Based on Dice

3) Based on playing Cards

4) Based on Marbles or balls

5) Miscellaneous

Important Questions:

1. Question A coin is thrown two times .what is the probability that at least one tail is obtained?

A) 3/4 B) 1/4 C) 1/3 D) 2/3 E) None of these

Answer :- A Sol:

Sample space = [TT, TH, HT,HH] Total number of ways = 2 × 2 = 4. Favourite Cases = 3 P (A) = 3/4

Tricks:- P (of getting at least one tail) = 1 – P (no head)⇒ 1 – 1/4 = 3/4

2. Question What is the probability of getting a numbered card when drawn from the pack of 52 cards?

A) 1/13 B) 1/9 C) 9/13 D) 11/13 E) None of these

Answer :- C Sol: Total Cards = 52. Numbered Cards = 9 (2,3,4,5,6,7,8,9,10) in each suit Numbered cards in four suit = 4 ×9 = 36 P (E) = 36/52 = 9/13

3.Question There are 7 purple clips and 5 brown clips. Two clips are selected one by one without replacement. Find the probability that the first is brown and the second is purple.

A) 1/35 B) 35/132 C) 1/132 D) 35/144 E) None of these

Answer :- B Sol:

P (B) × P (P) = (5/12) x (7/11) = 35/132

4.Question Find the probability of getting a sum of 8 when two dice are thrown?

A) 1/8 B) 1/5 C) 1/4 D) 5/36 E) 1/3

Answer 😀 Sol: Total number of ways = 6 × 6 = 36 ways. Favorable cases = (2 , 6) (6, 2) (3, 5) (5, 3) (4, 4) — 5 ways. P (A) = 5/36 = 5/36

5.Question Find the probability of an honour card when a card is drawn at random from the pack of 52 cards.

A) 4/13 B) 1/3 C) 5/12 D) 7/52 E) None of these

Answer :-A Sol: Honor cards = 4 (A, J, Q, K) in each suit Honor cards in 4 suit = 4 × 4 = 16 P (honor card) = 16/52 = 4/13

6. Question What is the probability of a face card when a card is drawn at random from the pack of 52 cards?

A) 1/13 B) 2/13 C) 3/13 D) 4/13 E) 5/13

Answer :-C Solution: face cards = 3 (J,Q,K) in each suit Face cards in 4 suits = 3 × 4 = 12 Cards. P (face Card) = 12/52 = 3/13

7.Question If two dice are rolled together then find the probability as getting at least one ‘3’?

A) 11/36 B) 1/12 C) 1/36 D) 13/25 E) 13/36

Answer :- A Sol: Total number of ways = 6 × 6 = 36. Probability of getting number ‘3′ at least one time = 1 – (Probability of getting no number 4) = 1 – (5/6) x (5/6) = 1 – 25/36 = 11/36

8. Question If a single six-sided die is rolled then find the probability of getting either 3 or 4.

A) 1/2 B) 1/3 C) 1/4 D) 2/3 E) 1/6

Answer:- B Solution:- Total outcomes = 6 Probability of getting a single number when rolled a die = 1/6 So, P(3) = 1/6 and P(4) = 1/6 Thus, the probability of getting either 3 or 4 = P(3)+P(4) = 1/6 + 1/6 = 1/3

9. Question A container contains 1 red, 3 black, 2 pink and 4 violet gems. If a single gem is chosen at random from the container, then find the probability that it is violet or black?

A) 1/10 B) 3/10 C) 7/10 D) 9/10 E) None of these

Answer :-C Sol :- Total gems =( 1 + 3 + 2 + 4 ) = 10 probability of getting a violet gem = 4/10 Probability of getting a black gem = 3/10 Now, P ( Violet or Black) = P(violet) + P(Black) = 4/10 + 3/10 = 7/10

10.Question A jar contains 63 balls ( 1,2,3,……., 63). Two balls are picked at random from the jar one after one and without any replacement. what is the probability that the sum of both balls drawn is even?

A) 5/21 B) 3/23 C) 5/63 D) 19/63 E) None of these

Answer :- A Sol. Total balls = 63 Total even balls = 31 ( 2 , 4 , 6,……., 62) Now the required probability =³¹C₂/63C₂ = (31!/2!29!)/(63!/2!61!) = (31 × 30/1× 2)/(63×62/1×2) = (31 × 30)/(63×62) = 30/63×2 = 5/21

11.Question There are 30 students in a class, 15 are boys and 15 are girls. In the final exam, 5 boys and 4 girls made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an ‘A grade student?

A) 1/4 B) 3/10 C) 1/3 D) 2/3 E) None of these

Answer:- D Sol:

Here, the total number of boys = 15 and the total number of girls = 15

Also, girls getting A grade = 4 and boys getting an A grade = 5 Probability of choosing a girl = 15/30

Probability of choosing A grade student= 9/30

Now, an A-grade student chosen can be a girl. So the probability of choosing it = 4/30

Required probability of choosing a girl or an A grade student = 15/30 + 9/30 – 4/30 = 1/2 + 3/10 – 2/15 = 2/3 12. Question What is the probability when a card is drawn at random from a deck of 52 cards is either an ace or a club?

A) 2/13 B) 3/13 C) 4/13 D) 5/23 E) None of these

Answer:- C Sol: There are 4 aces in a pack, 13 club cards and 1 ace of club card.

Now, the probability of getting an ace = 4/52

Probability of getting a club = 13/52

Probability of getting an ace of club = 1/52

Required probability of getting an ace or a club

= 4/52 + 13/52 – 1/52 = 16/52 = 4/13

13. Question One card is drawn from a deck of 52 cards well shuffling. Calculate the probability that the card will not be a king.

A) 12/13 B) 3/13 C) 7/13 D) 5/23 E) None of these

Solution:

Well-shuffling ensures equally likely outcomes. Total king of a deck = 4

The number of favourable outcomes F= 52 – 4 = 48

The number of possible outcomes = 52

Therefore, the required probability

= 48/52 = 12/13

14.Question If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, find the value of P(A|B).

A) 1/9 B) 2/9 C) 3/9 D) 4/9 E) None of these

Answer :- D Solution:

P(A|B) = P(A∩B)/P(B) = (4/13)/(9/13) = 4/9.

15. Question A one rupee coin and a two rupee coin are tossed once, then calculate a sample space.

The outcomes are either Head (H) or tail(T).

Now,heads on both coins = (H,H) = HH

Tails on both coins = ( T, T) = TT

Probability of head on one rupee coin and Tail on the two rupee coins = (H, T) = HT

And Tail on one rupee coin and Head on the two rupee coin = (T, H) = TH

Thus, the sample space ,S = [HH, HT, TH, TT]

16. Question There are 20 tickets numbered 1 to 20. These tickets are mixed up and then a ticket is drawn at random. Find the probability that the ticket drawn has a number which is a multiple of 4 or 5?

A) 1/4 B) 2/13 C) 8/15 D) 9/20 E) None of these

Here, S = {1, 2, 3, 4, …., 19, 20} = 20

Let E = event of getting a multiple of 4 or 5 = {4, 8 , 12, 16, 20, 5, 10, 15, 20} = 9

Required probability = favourable outcomes/total outcomes = 9/20

Direction ( 17 – 19):- In a school the total number of students is 300, 95 students like chicken only, 120 students like fish only, 80 students like mutton only and 5 students do not like anything above. If randomly one student is chosen, find the probability that

17) The student likes mutton.

18 ) he likes either chicken or mutton

19 ) he likes neither fish nor mutton.

Solution( 17-19):-

The total number of favourable outcomes = 300 (Since there are 300 students altogether).

The number of times a chicken liker is chosen = 95 (Since 95 students like chicken).

The number of times a fish liker is chosen = 120.

The number of times a mutton liker is chosen = 80.

The number of times a student is chosen who likes none of these = 5.

17. Question Find the probability that the student like mutton?

A) 3/10 B) 4/15 C) 1/10 D) 1/15 E) None of these

Answer:- B Solution:-

Therefore, the probability of getting a student who likes mutton

= 80/300 = 4/15

18. Question What is the probability that the student likes either chicken or mutton?

A) 7/12 B) 5/12 C) 3/4 D) 1/12 E) None of these

Answer:- A Solution:-

The probability of getting a student who likes either chicken or mutton = (95+80)/300 = 175/300 = 7/12

19. Question Find the probability that the student likes neither fish nor mutton.

A) 1/2 B) 1/5 C) 1/3 D) 1/4 E) 1/6

Answer:- C Solution:- The probability of getting a student who likes neither fish nor mutton = (300–120−80)/300 = 100/300 = 1/3

Direction ( 20-22):- A box contains 90 number plates numbered 1 to 90. If one number plate is drawn at random from the box then find out the probability that

20) The number is a two-digit number

21) The number is a perfect square

22) The number is a multiply of 5

20. Question Find the probability that the number is a two-digit number.

A) 1/9 B) 1/10 C) 9/10 D) 7/10 E) None of these

Answer:-C Solution : Total possible outcomes = 90 (Since the number plates are numbered from 1 to 90).

Number of favourable outcomes = 90 – 9 = 81 ( here, except 1 to 9, other numbers are two-digit number.)

Thus required probability = Number of Favourable Outcomes /Total Number of Possible Outcomes = 81/90 = 9/10.

21. Question What is the probability that the number is a perfect square?

A) 1/9 B) 1/10 C) 9/10 D) 1/7 E) None of these

Answer:- B Solution:- Total possible outcomes = 90. Number of favourable outcomes = 9 [here 1, 4, 9, 16, 25, 36, 49, 64 and 81 are the perfect squares] Thus the required probability = 9/90 =1/10

22.Question Find the probability that the number is a multiple of 5.

A) 1/5 B) 1/6 C) 1/10 D) 1/8 E) 9/10

Answer:- A Solution:- Total possible outcomes = 90. Number of favourable outcomes = 18 (here, 5 × 1, 5 × 2, 5 × 3, …., 5 × 18 are multiple of 5).

Thus, the required probability= 18/90 =1/5

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Normal Distribution Formula

What is a Normal Distribution?

The normal distribution is the most significant probability distribution in statistics as it is suitable for various natural phenomena such as heights, measurement of errors, blood pressure, and IQ scores follow the normal distribution. The other names for the normal distribution are Gaussian distribution and the bell curve.

The normal distribution is a probability distribution that outlines how the values of a variable are distributed. Most of the observations clusters around the central peak and probabilities for values far away from mean fall off equally in both the directions in a normal distribution. Extreme values in both the tails of the distribution are uniformly unexpected.

The spread of the normal distribution is managed by the standard deviation . The smaller the standard deviation value in a normal distribution formula, the more concentrated the data.

The normal probability distribution formula is given as:

\[P (x) = \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-\frac{(x - \mu)^{2}}{2 \sigma^{2}}}\]

In the above normal probability distribution formula.

μ is the mean of the data.

σ is the standard deviation of data.

x is the normal random variable.

If (μ) = 0 and standard normal deviation is equal to 1, then distribution is said to be a normal distribution.

What is the Probability Distribution Formula?

There are generally two types of probability distribution formulas in a probability distribution. The two types of probability distribution formulas are normal probability distribution formulas (also known as the Gaussian distribution formulas) and binomial distribution formulas.

The value that assigns a probability to each outcome of a random experiment is considered as a probability distribution equation. In other words, it assigns the probability for each value of the random variable.

Probability Distribution Formula

The two types of probability distribution formulas are:

The normal probability distribution formula.

The binomial Probability Distribution Formula.

Normal Probability Distribution Formula

The normal distribution is also known as the Gaussian distribution and it denotes the equation or graph which are bell-shaped.

The normal probability distribution formula is given by:

Binomial Probability Distribution Formula

These are the probabilities that appear when the event consists of n repeated trials and the results of each trial may or may not appear.

The binomial probability distribution formula is stated below:

P ( r out of n) = n!/ r!(n-r)!

p r (1-p) n-r = n C r

p r (1-p) n-r

In the above binomial distribution formula,

N is the total number of events

r is the total number of successful events

p is the probability of success on each trial.

nC r = n!/r!(n-r)!

1- p = probability of failure.

What is a Lognormal Distribution?

In probability theory, a log-normal distribution is considered as the continuous probability distribution of a random variable whose logarithm follows the pattern of normal distribution. It represents phenomena whose relative growth rate is independent of the size, which is true for most of the natural phenomena including the size of tissue and blood pressure, income distribution and then the length of the chess.

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Lognormal Distribution Formula

Some of the lognormal distribution formulas are given below:

The lognormal distribution formula for mean is given as

m = eμ + σ² /2

Which implies that μ can be calculated from m:

m = In m – 1/2 σ²

The above both equations are derived from the mean of the normal distribution.

The lognormal distribution formula for the median is given as

Med [X] = e μ

The above equation is derived by placing the cumulative distribution values equal to 0.5 and solving the resulting equation

The lognormal distribution formula for mode is given as

Mode [X] = e μ - σ²

The above equation is derived by placing the probability distribution function values equal to 0 as the mode denotes the global maximum of distribution.

The lognormal distribution formula for variance is given as:

Var [X] = (e σ² -1) e 2μ + σ² ,

Which can also be represented as (e σ² -1) m 2 , where m denotes the mean of the distribution.

Gaussian Distribution Formula

Gaussian distribution is a common continuous probability distribution. The Gaussian Distribution Formulas are significant in statistics and are primarily used in natural and social science to denote real-valued random variables.

The probability density function formula for Gaussian distribution formula is given as:

\[f( x,\mu, \sigma) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x - \mu)^{2}}{2 \sigma^{2}}}\]

In the above Gaussian Distribution Formula,

Standard Normal Distribution

The standard normal distribution is a type of normal distribution. It mostly appears when a normal random variable has a mean value equal to 0 and value of standard deviation is equal to 1. The random variable of a standard normal distribution is considered as a standard score or z-score. Each normal random variable such as X can easily be converted into a z-score using the normal distribution z formula.

\[z = \frac{(X - \mu)}{\sigma}\]

In the above normal distribution z formula,

X is a normal random variable.

Μ is mean of data.

σ is the standard deviation of the data.

Solved Examples

1. Calculate the probability of normal distribution with the population mean 2, standard deviation 3 or random variable 5.

Mean = μ = 2

Standard Deviation = σ = 3

We will solve the questions with the help of the above normal probability distribution formula:

\[P (x) = \frac{1}{\sqrt{2 \times 3.14 \times 3^{2}}} e^{-\frac{(5 - 2)^{2}}{2 \times 3^{2}}}\]

= \[\frac{1}{\sqrt{56.52}} e^{-\frac{9}{2 \times 9}}\]

= \[\frac{1}{7.518} e^{-\frac{1}{2}}\]

= \[\frac{0.6065}{7.518}\]

= \[0.0807\]

2. An average electric bulb lasts for 300 days with a standard deviation of 50 days. Assume that bulb life is normally distributed, what is the probability that the electric bulb will last at most 365 days.

Solution :

Given, Mean score = 300 days

Standard deviation = 50 days

We need to determine the cumulative probability that bulb life is less than or equals to 365 days.

We have the following information:

The normal random variable value is given as 365 days.

The mean is equivalent to 300 days.

The standard deviation is equivalent to 50 days.

Hene, the answer of the question is P (x < 365) = 0.90.

It states that there is a 90% probability that an electric bulb will burn out within 365 days.

1. The shape of the normal curve is

2. The area under a standard normal curve is

Not defined

FAQs on Normal Distribution Formula

1. What are the Common Properties for all Forms of Normal Distribution?

Ans: In Spite of different shapes, all forms of normal distribution have the following characteristics properties.

The mean, median, and mode values are equal.

Half of the population is less than mean and half is greater than mean.

They are all symmetric. The normal distribution cannot model skewed distribution.

The Empirical rule regulates the proportion of values that lies within a certain distance from the mean.

The total area under the curve is 1.

The bell curve is symmetric at the center ( i.e. around the mean, μ).

2. Why is the Normal Distribution Important?

Ans: The normal distribution uses a continuous probability distribution that is symmetrical on a shape on both sides of the mean, so the right side of the image is a mirror image of the left side. The area under the normal distribution curve denotes probability and the total area under the curve is equal to one. The normal distribution is the most important probability distribution in statistics because various continuous data and psychology in nature exhibit this bell -shaped curve when collected and graphed.

For example, if we randomly sample 100 individuals, we would expect to see a normal distribution curve of various continuous variables such as IQ, height, weight, and blood pressure.

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