**Binomial distribution practice problems**

**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.

**Show Solution**

**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.

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

**Show Solution**

**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)

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

= 40C_{14 }* (0.2)^{14 }* (0.8)^{40-14 }= 40C_{14 }* (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 }* (0.8)^{0 }* (0.2)^{40-0 }= 40C_{0 }* (0.8)^{0 }* (0.2)^{40 }is approximately 0

P(y= 1) = 40C_{1 }* (0.8)^{1 }* (0.2)^{40-1 }= 40C_{1 }* (0.8)^{1 }* (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)]

**Show Solution**

**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)

Hence, the answer is a)

[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?

**Show Solution**

**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)

**P(S = 0) + P(S = 1) + P(S = 2) +……..P(S = 7) = 1 ****Sum of all probabilities of all possibilities**

### P(S = 0) = 7C0 * (0.13)^0 * (0.87)^7 = 0.377

### Probability of at least one defective item out of 7 items selected = 1 – P(S = 0)

### = 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. **

**If 60 trials are made,**

a) What is the expected number of successes?

b) Calculate the standard deviation of number of successes.

**Show Solution**

**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?

**Show Solution**

**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?

**Show Solution**

**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:

- n*p > 10 {Here n*p = 70}
- 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?

**Show Solution**

**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?

**Show Solution**

**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.

**Show Solution**

**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?

**Show Solution**

**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**