# Binomial distribution problem 1

Question (Binomial distribution problems): Expected value of a binomial random variable X is 10 and the probability of failure is 0.9. Calculate the standard deviation of X.

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## Solution:

For a binomial random variable, mean and variance are given by below formula.

Mean = np = 10

p = 1 – 0.9 = 0.1

n = 10/0.1 = 100

Variance = n*p*q = 100 * 0.1 * 0.9 = 9

Standard deviation of X = 3

# Binomial distribution problem 2

Question (Binomial distribution problems): Sam has not prepared for the upcoming exam. Exam is completely based on multiple choice questions with 5 options for each question and only one correct answer. There are 30 questions in the exam and Sam attempted all questions betting on his luck.

1. What is the probability that exactly 17 questions are correct?
2. What is the probability that at least 2 questions are wrong?

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## Solution:

Number of correct answers that Sam gets in the exam follows binomial distribution (Read the conditions under which a distribution becomes binomial here – Binomial, negative binomial and geometric distributions)

Probability of correct answer p = 0.2 (1 out 5 options is correct)

So, Probability of wrong answer q = 0.8 (4 out of 5 options)

1. P( x = 14) where x is number of correct answers

= 40C14 * (0.2)14 * (0.8)40-14 = 40C14 * (0.2)14 * (0.8)26 = 0.01149

### Concept: Formula for Binomial distribution is given below

2. P(at least 2 questions are wrong) is required

P(y = 0) + P(y = 1) + P(y = 2) ……..+ P(y=40) = 1 [as Number of wrong answers could vary between 0 and 40 only]

Where y is number of wrong answers

P (at least 2 questions are wrong) = P(y = 2) + P(y = 3)……..+ P(y=40)

= 1 – {P(y = 0) + P(y = 1)}

Here y is number of wrong answers which is considered as success (Success is what we require)

Probability of wrong answer/Success p = 0.8 (4 out of 5 options)

Probability of correct answer/Failure q = 0.2 (1 out 5 options is correct)

P(y= 0) = 40C* (0.8)* (0.2)40-0 = 40C* (0.8)* (0.2)40 is approximately 0

P(y= 1) = 40C* (0.8)* (0.2)40-1 = 40C* (0.8)* (0.2)39 is approximately 0

P (at least 2 questions are wrong) is required = 1 – {P(y = 0) + P(y = 1)} is approximately

= 1-{0+0} = 1

Binomial distribution practice tests

# Binomial distribution problem 3

Question (Binomial distribution problems): Below are the set of binomial distributions for n = 20 (20 trials) and for varying probability of success. Choose the right match from the below!

a) [A – Binom(20, 0.7); B – Binom(20, 0.3); C – Binom(20, 0.5); D – Binom(20, 0.98)]

b) [A – Binom(20, 0.1); B – Binom(20, 0.8); C – Binom(20, 0.5); D – Binom(20, 0.2)]

c) [A – Binom(20, 0.7); B – Binom(20, 0.3); C – Binom(20, 0.8); D – Binom(20, 0.1)]

d) [A – Binom(20, 0.3); B – Binom(20, 0.7); C – Binom(20, 0.5); D – Binom(20, 0.98)]

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Solution:

For P = 0.5, binomial distribution appears to be close to normal distribution,

For P < 0.5 binomial distribution is right skewed (Long tail to the right); For P> 0.5 distribution is left skewed (Long tail to the left)

[A – Binom(20, 0.7); B – Binom(20, 0.3); C – Binom(20, 0.5); D – Binom(20, 0.98)]

# Binomial distribution problem 4

Question (Binomial distribution problems): It is known that 13% of all items produced by a machine are defective. If 7 items are selected at random,

a) What is the probability that exactly 2 items are defective?

b) What is the probability that at least one the items selected is defective?

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## Solution

### Step 1: Identify the type of distribution

Number of defective items selected out of n trials follows binomial distribution

(Read conditions for binomial distribution – Binomial, negative binomial and geometric distributions)

### Step 2: Define parameters

Success – Selecting a defective item

Failure – Selecting a non-defective item

P(Success) = 0.13

P(failure) = 0.87

## a)

Probability that exactly 3 items are defective out of the 7 items

P(S = 2) = 7C2 * (0.13)^2 * (0.87)^5 = 0.177

## Alternatively, we could define success as

Success – Selecting a non-defective item

Failure – Selecting a defective item

P(Success) = 0.87

P(failure) = 0.13

Selecting exactly 2 defective items means selecting exactly 5 non-defective items

P(S = 5) = 7C5 * (0.87)^5 * (0.13)^2 = 0.177

## b)

Probability of selecting at least one defective item is P(S >=1)

### = 1 – 0.377 = 0.623

Binomial distribution practice tests

# Binomial distribution problem 5

Question (Binomial distribution problems): Let X be a binomial random variable with probability of success as 0.3.

a) What is the expected number of successes?

b) Calculate the standard deviation of number of successes.

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Solution:

a) Expected number of success is same as mean.

Mean and variance of a binomial distribution is given by below formulas

Mean or expected value = n*p = 60*0.3 = 18

Variance = n*p*q = 60*0.3*0.7 = 126

Standard deviation = SQRT(Variance) = SQRT(126) = 11.22

Binomial distribution practice tests

# Binomial distribution problem 6

Question (Binomial distribution problems): It is known that 15% of all the items produced from a factory are defective.

a) What is the expected number of items to be drawn to get a non-defective item?

b) Calculate the expected number of items to be drawn to get a defective item?

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## Solution:

Probability of drawing either defective item or a non-defective item is constant. Number of successes X follows binomial distribution,.

Expected value for binomial distribution is given by below formula

n * p where n is number of trials and p is probability of success.

a)

Expected value of drawing non-defective items = n * P(non-defective)

P(non – defective) = 0.85

n * 0.85 = 1 { To get one non-defective item}

n = 1/0.85 = 1.176

b)

Expected value of drawing defective items = n * P(defective)

P(defective) = 0.15

n * 0.15 = 1 { To get one non-defective item}

n = 1/0.15 = 6.66

Binomial distribution practice tests

# Binomial distribution problem 7

Question (Binomial distribution problems): An urn contains several marbles. It is known that 35% of all the marbles in the urn are blue. If a sample of 200 marbles are selected are random, what is the probability of getting at-least 80 blue marbles?

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## Solution:

Selection of blue marbles follows binomial distribution. Required probability = P(S = 80) + P(S = 81) + P(S = 82) +….P(S = 200)

Calculation of each of the above probabilities is very tedious. In such cases, normal approximation to binomial distribution will work if  the conditions are met.

Conditions for normal approximation:

1. n*p  > 10 {Here n*p = 70}
2. n*(1-p) >10 {Here n*(1-p) = 130}

This distribution could be considered as normal with mean as np and variance as np(1-p)

Mean = 70 and Variance = 45.5

SD = 6.74

P(S>=80) = P(Z >= ((80 – 70)/6.74)

Required probability = P(Z > 1.48) = 0.069 { Normal distribution tables and areas ( z tables)}

# Binomial distribution problem 8

Question (Binomial distribution problems): Kate is playing a game with her friend.  She wins the game if she gets at least 4 heads out of 6 times when a fair coin is tossed. What is the probability that Kate loses the game?

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## Solution:

Probability that kate loses the game = 1 – Probability she wins the game

Required probability = 1 – P(Kate wins)

p(Kate wins) = P(X = 4) + P(X = 5) + P(X = 6)

probability of success p = 0.5 (Fair coin)

P(X = 4) = 6C4 * (0.5) ^ 4 * (0.5) ^ 2 = 0.234

P(X = 5) = 0.093

P(X = 6) = 0.015

P(Kate wins) = 0.23 + 0.09 + 0.015 = 0.342

P(Kate loses) = 1 – 0.342 = 0.658 ~ 0.65

Binomial distribution practice tests

# Binomial distribution problem 9

Question (Binomial distribution problems): A fair dice is rolled 5 times. What is the probability that a number greater than 4 occurs at least 3 times?

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## Solution:

Probability that a number greater than 4 occurs in each trial = 2/6 = 1/3

p = 1/3;  q = 2/3

Let X be the number of successes

Required probability = p(X = 3) + p(X = 4) + P(X = 5)

p(X= 3)  =  5c3 * (1/3) ^ 3 * (2/3) ^ 2 = 0.165

p(X = 4) = 0.041

P(X = 5) = 0.004

Required probability = 0.21

Binomial distribution practice tests

# Binomial distribution problem 10

Question (Binomial distribution problems): It is a known that 70% of the electric bulbs procured from a particular manufacturer lasts for at least 240 days before they fail. If a sample of 6 bulbs are drawn at random, find the probability that at least 5 bulbs fail before 240 days.

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## Solution:

Probability that at least 5 bulbs fail = P(X = 5) + P(X = 6)

Here we are concerned about bulb failure which is ‘success’.

p = 0.3; q = 0.7

P(X = 5) = 6C5*(0.3) ^ 5 * (0.7)^1 = 0.0102

P(X = 6) = 0.0007

Required probability = 0.0109

Binomial distribution practice tests

# Binomial distribution problem 11

Question (Binomial distribution problems): A fair dice is rolled 6 times, what is the probability that a number between 3 and 6 (including both) occurs at least 3 times and no more than 5 times?

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## Solution:

In this case, ‘Success’ is getting a number between 3 and 6 (Including both) in each trail.

p(success) = p = 4/6 = 2/3

q = 1/3

Required probability = P(X = 3) + P(X = 4) + P(X = 5)

P(X = 3) = 6C3 * (2/3) ^ 3 * (1/3) ^3 = 0.219

P(X = 4) = 0.329

P(X = 5) = 0.263

Required probability = 0.811

Binomial distribution practice tests