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## 50+ Solved Probability Distribution MCQs – Statistics

## Probability Distribution MCQs

If 0 A. 1 B. 1/? C. ? D. None of these View Answer B. 1/?

Read Also>> Statistical Inference MCQs

Mean of Gamma distribution is. A. (a-1)ß B. aß C. (a-1)ß2 D. aß2 View Answer B. aß

If f(x) = 1/2 e^(-|x| ), -8=x=8 then its mean deviation is. A. 4 B. 2 C. 1 D. 0 View Answer C. 1

Range of Cauchy distribution is. A. -8 to 8 B. 0 to ? C. 0 to 8 D. -1 to +1 View Answer A. -8 to 8

Mean of ?2 distribution is. A. 2n B. n-2 C. n D. 8n View Answer C. n

Mode of ?2 distribution is. A. 2n B. n-2 C. n D. 8n View Answer B. n-2

?^2?normality if. A. n?8 B. n?0 C. n?10 D. n?1 View Answer A. n?8

If n ?8 then t ?. A. Normal B. Gamma C. Beta D. Laplace View Answer A. Normal

If x ~ F(n_1,n_2) then ……..~ F(n_2,n_1 ). A. x2 B. 1/x C. ?2 D. None of these View Answer B. 1/x

Mean of F-distribution exists for. A. n2> 2 B. n2> 4 C. n2> 1 D. None of these View Answer A. n2> 2

If y1 = y2 = y3 then first order statistics is. A. y3 B. y2 C. y1 D. None of these View Answer C. y1

If f(x) = e-x, 0 A. e-x B. 1- e-x C. e-x -1 D. None of these View Answer D. None of these

If y1 = y2 = y3 then median is. A. y3 B. y2 C. y1 D. None of these View Answer A. y3

Which is order statistics. A. A. M B. Median C. G. M D. H. M View Answer B. Median

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## Probability Distribution | Formula, Types, & Examples

Published on June 9, 2022 by Shaun Turney . Revised on November 10, 2022.

## Table of contents

A probability distribution is an idealized frequency distribution .

The higher the probability of a value, the higher its frequency in a sample.

The farmer weighs 100 random eggs and describes their frequency distribution using a histogram:

- Random variables are usually denoted by X.
- The ~ (tilde) symbol means “follows the distribution.”
- The distribution is denoted by a capital letter (usually the first letter of the distribution’s name), followed by brackets that contain the distribution’s parameters .

There are two types of probability distributions:

## Probability tables

A probability table is composed of two columns:

## Probability mass functions

A probability mass function can be represented as an equation or as a graph.

The probability mass function of the distribution is given by the formula:

This probability mass function can also be represented as a graph:

## Common discrete probability distributions

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A continuous probability distribution is the probability distribution of a continuous variable .

## Probability density functions

A probability density function can be represented as an equation or as a graph.

## Common continuous probability distributions

- If you have a formula describing the distribution, such as a probability density function, the expected value is usually given by the µ parameter. If there’s no µ parameter, the expected value can be calculated from the other parameters using equations that are specific to each distribution.
- If you have a sample , then the mean of the sample is an estimate of the expected value of the population’s probability distribution. The larger the sample size, the better the estimate will be.
- If you have a probability table , you can calculate the expected value by multiplying each possible outcome by its probability, and then summing these values.

What is the expected value of robin eggs per nest?

Multiply each possible outcome by its probability:

The standard deviation of a distribution is a measure of its variability. It’s often written as σ.

- If you have a formula describing the distribution, such as a probability density function, the standard deviation is sometimes given by the σ parameter. If there’s no σ parameter, the standard deviation can often be calculated from other parameters using formulas that are specific to each distribution.
- If you have a sample , the standard deviation of the sample is an estimate of the standard deviation of the population’s probability distribution. The larger the sample size, the better the estimate will be.
- If you have a probability table , you can calculate the standard deviation by calculating the deviation between each value and the expected value, squaring it, multiplying it by its probability, and then summing the values and taking the square root.

Square the values and multiply them by their probability:

Sum the values and take the square root:

Probability is the relative frequency over an infinite number of trials.

## Cite this Scribbr article

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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## Shaun Turney

## Probability Questions

FACTS AND FORMULAE FOR PROBABILITY QUESTIONS

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

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.

iii. A pack of cards has 52 cards.

- It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds.
- Cards of spades and clubs are black cards.
- Cards of hearts and diamonds are red cards.

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.

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 )

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

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

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

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

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}.

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

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

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

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.

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

- Compute the probability function for a certain probability distribution
- Apply the probability distribution to solve a problem.

The essential knowledge for all these questions is the following:

- “The first defective product”: this is our event, and we want to know when it happens. We define our variable as X = “The number of the first defective trial”
- “the fifth trial”: this gives us the range of values we want consider.

This is all we need to write our code.

This should be broken down for clarity:

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

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

The code to these two questions then is:

- Scored higher than 80 ?
- Passed the test (i.e., scored at least 60 )?
- Failed the test (i.e., scored less than 60 )?

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## 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.

## Table of contents

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

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

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

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

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;

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)

Calculation of Probability of an event

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

- So, the probability of selecting 1 woman = no of the possibility of selecting 1 women / total possibilities
- Probability of selecting 2 women =no of the possibility of selecting 2 women / total possibilities

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

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

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

Similarly, Probability of selecting only 1 damage light = [P(G) X P(G) X P(D)] X 3

Similarly, Probability of selecting 2 damage lights = [P(G) X P(D) X P(D)] X 3

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

- Normal Distribution
- Exponential Distribution Exponential Distribution Exponential distribution refers to the continuous and constant probability distribution which is actually used to model the time period that a person needs to wait before the given event happens. This distribution is a continuous counterpart of a geometric distribution that is instead distinct. read more
- Poisson Distribution Meaning Poisson Distribution Meaning Poisson distribution refers to the process of determining the probability of events repeating within a specific timeframe. read more
- Lognormal Distribution Excel Lognormal Distribution Excel Lognormal Distribution finds out the distribution of a variable whose logarithm is normally distributed. Excel has an inbuilt function to calculate the lognormal distribution. read more

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## Probability Questions And Answers

## Probability Formula

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

## Solved Probability Examples

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

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

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

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

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 other coloured balls = 5

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

b] either cricket or volleyball sport

c] neither football nor volleyball

Total number of students = 300

5 students don’t play any games.

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

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

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.

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

## Meaning and definition of Probability

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

- Tossing a coin with the head up
- Drawing a red pen from a pack of different coloured pens
- Drawing a card from a deck of 52 cards etc.

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.

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.

Similarly when two coins are tossed, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

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

## Basic formula of probability

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

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

Total no. of possible outcomes = 6

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

## Compound probability

## Formula for compound probability

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:

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)

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)

## Independent and Dependent 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)}.

Here, total number of pens = 9

## Dependent Events

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

Let’s consider another example:

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

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

Consider the following example:

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’).

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

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

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

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)

## Probability Example 2

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

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)}.

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)

## Probability Quiz: Sample probability questions for practice

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

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

The math here is totally wrong

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

Hi Last question must be 212/216 right ?

Tell me the way u did that sol. Plzz…

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

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

copying the solution offerred by @ diriba

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

This will give you P(Bna)= 0.3.

This will give you 3/7 as the answer.

please help me in solving this

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

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

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%

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

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## Free Mathematics Tutorials

Poisson probability distribution examples and questions.

## Poisson Process Examples and Formula

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

- My computer crashes on average once every 4 months.
- Hospital emergencies receive on average 5 very serious cases every 24 hours.
- The number of cars passing through a point, on a small road, is on average 4 cars every 30 minutes.
- I receive on average 10 e-mails every 2 hours.
- Customers make on average 10 calls every hour to the customer help center

## Compare Binomial and Poisson Distributions

## More References and links

## By Gkseries more articles

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

Answer: Only if random variables exhibit statistical independency

Answer: P (G) = 1 – P (H)

## DOWNLOAD CURRENT AFFAIRS PDF FROM APP

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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## Current Affairs MCQs

## Binomial Distribution Questions

- n is finite and defined.
- Each trial has only two possible outcomes: success and failure.
- The result of each trial is independent of other trials.
- The probability of success and failure remains the same in each trial.

## Bernoulli’s Theorem for Binomial Distribution

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

- Sum of all probabilities in the distribution sums up to 1
- Probability of success in all n trials is p n
- Probability of failure in all n trials is (1 – p) n = q n
- Probability of success in at least one trial = P(X ≥ 1) = 1 – P(X = 0) = 1 – q n .
- Probability of at least r successes = P(X ≥ r) = ∑ k n C k p k q n – k (k = r, r + 1,…, n)
- Probability of at most r successes = P(X ≤ r) = ∑ k n C k p k q n – k (k = 0, 1, …, r)
- If in n trials, the experiment is repeated N times, the expected frequencies are N.P(r) for r = 0, 1, 2, 3, …, n.

Learn more about binomial distribution .

## Binomial Distribution Questions with Solutions

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

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.

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

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

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

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

The required binomial distribution of X is:

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

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

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

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

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

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

∴ The required probability distribution is:

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.

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)

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

= 3 × (2/7) × (25/49) = 150/343

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

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

The required probability distribution:

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)

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.

= 5 C 2 p 2 q 3 × 1 C 1 p 1 q 1 – 1

= 10 × (⅙) 2 × (⅚) 3 × 1 × (⅙)

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

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

Thus, he must answer at least 4 questions.

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

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

The probability distribution for getting the tail is:

Also try: Binomial Distribution Calculator

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:

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

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%?

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## Tricks To Solve Probability Questions

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.

Types of questions asked in the competitive exam:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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]

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

18 ) he likes either chicken or mutton

19 ) he likes neither fish nor mutton.

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

Therefore, the probability of getting a student who likes mutton

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

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

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

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

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

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

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## What is a Normal Distribution?

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.

σ is the standard deviation of data.

x is the normal random variable.

## What is the Probability Distribution Formula?

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 probability distribution formula is given by:

## Binomial Probability Distribution Formula

The binomial probability distribution formula is stated below:

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

1- p = probability of failure.

## What is a Lognormal Distribution?

## Lognormal Distribution Formula

Some of the lognormal distribution formulas are given below:

The lognormal distribution formula for mean is given as

Which implies that μ can be calculated from m:

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

The lognormal distribution formula for the median is given as

The lognormal distribution formula for mode is given as

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

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

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

In the above normal distribution z formula,

X is a normal random variable.

σ is the standard deviation of the data.

## Solved Examples

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

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

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

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

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

## FAQs on Normal Distribution Formula

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

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 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?

## IMAGES

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Probability Distribution Questions and Answers Test your understanding with practice problems and step-by-step solutions. Browse through all study tools. Questions and Answers ( 4,651 ) It is...

Probability Distribution MCQs Mean of continuous uniform (Rectangular) distribution is. A. (ß+a)/2 B. (ß-a)/2 C. (ß-a)/4 D.? (ß-a)?^2/12 View Answer Mean deviation of continuous uniform (Rectangular) distribution is. A. (ß+a)/2 B. (ß-a)/2 C. (ß-a)/4 D. ? (ß-a)?^2/12 View Answer If 0 A. 1 B. 1/? C. ? D. None of these View Answer

The probability mass function of the distribution is given by the formula: Where: is the probability that a person has exactly sweaters is the mean number of sweaters per person (, in this case) is Euler's constant (approximately 2.718) This probability mass function can also be represented as a graph:

Probability Questions Popular Latest Rated Important Formulae Q: A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour. View Answer Report Error Discuss Filed Under: Probability - Quantitative Aptitude - Arithmetic Ability 426 77116 Q:

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, . This observation is enough to solve these three questions.

Calculation of probability of an event can be done as follows, Using the Formula, Probability of selecting 0 Head = No of Possibility of Event / No of Total Possibility = 1/4 Probability of an event will be - =1/4 Probability of selecting 1 Head = No of Possibility of Event / No of Total Possibility = 2/4 = 1/2

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}

KTU SOLVED QUESTION PAPERS Page Title Probability,distribution (MA202) Solved Question Paper 1 This Question PAPERS was contributed by ABHIRAMI SUDHARSHAN Anyone can contribute notes to this platform making this more methodically efficient and updated Student @ KTU Contribute here

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

Solution to Example 2. a) The average λ = 1 every 4 months. Hence the probability that my computer does not crashes in a period of 4 month is written as P(X = 0) and given by. P(X = 0) = e − λλx x! = e − 110 0! = 0.36787. b) The average λ = 1 every 4 months.

Question: In the probability distribution to the right, the random variable \( X \) represents the number of hits a baseball player obtained in a game over the course of a season. Complete parts (a) through ( \( f \) ) below. (a) Verify that this is a discrete probability distribution. This is a discrete probability distribution because between and inclusive, and the

SOLVED: Probability distribution Question Answered step-by-step Probability distribution Video Answer: Solved by verified expert Best Match Question: Probability distribution Recommended Videos 08:25 a random variable x can assume five values 0,1,2,3,4....... 01:26 Probability distribution 05:57 Probability of normal distribution 05:54

Question 6 (15 points) Using the probability distribution from Question 5, find the probability of 2 or 3 O.8 O.4 O . 2 0 .6 Question 7 (15 points) using the probability distribution from problem 5 find the probability of not getting a 3.

Which of the following probability distributions can be used to calculate the student's chance of getting at least 20 questions right? A Uniform distribution B Exponential distribution C Poisson distribution D Binomial distribution 17 Which of the following statements is/are true regarding the normal distribution curve?

solution:from the given data:E (x)=∑xp (x)E (x)=−250 … View the full answer Transcribed image text: QUESTION 5 The probability distribution shown above is for the MRA Company's projective profits ( x = profit in $1,000 s) for the first year of operation (negative values denote a loss).

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. Solution: Let X be the random variable of getting six. Then X can be 0, 1, 2, 3. Here, n = 3

The probability distribution function is essential to the probability density function. This function is extremely helpful because it apprises us of the probability of an affair that will appear in a given intermission. P (a<x<b) = ∫ba f (x)dx = (1/σ√2π)e[- (x - μ)²/2σ²]dx. Where.

The probability mass function (pmf) can be described as the probability that a discrete random variable X will be exactly equal to some value x. The geometric distribution pmf formula is as follows: P (X = x) = (1 - p) x - 1 p where, 0 < p ≤ 1 Geometric Distribution CDF

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]

Solved Examples 1. Calculate the probability of normal distribution with the population mean 2, standard deviation 3 or random variable 5. Solution: x = 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) = 1 2 π σ 2 e − ( x − μ) 2 2 σ 2