Practice Questions on Binomial Distribution with Solution

Q1: If a coin is tossed 3 times, then find the probability of getting exactly two heads.

Solution:

X be the random variable for number of heads

The formula for the required probability is given by:

P (X = x) = ⁿC_x p^x q^n-x

Here, n = 3, x = 2, p =1/2, q = 1/2

⇒ P (X = 2) = ³C₂ (1/2)² (1/2)^3-2

⇒ P (X = 2) = 3 (1/2)³

⇒ P(X = 2) = 3/8

Q2: A pair of dice is rolled 5 times. If getting the product of 6 is considered as a success. Find the probability of getting at least 4 successes.

Solution:

The formula for the above probability is given by:

P (X = x) = ⁿC_x p^x q^n-x

Here, n = 5

p is the probability of getting product 6

p = 4 /36 = 1/9

q = 1 – p = 8/9

⇒ P(X ≥ 4) = P(X = 4) + P(X = 5)

⇒ P(X ≥ 4) = ⁵C₄ (1/9)⁴ (8/9)^5-4+ ⁵C₅ (1/9)⁵ (8/9)^5-5

⇒ P(X ≥ 4) = 5(1/9)⁴ (8/9) + (1/9)⁵

Q3: If the number of trials of a certain binomial distribution is 225 and the probability of success is 0.36. Find the standard deviation.

Solution:

The formula of the standard deviation of binomial distribution is given by:

σ = √(npq)

Here, n = 225, p = 0.36 and q = 0.64

⇒ σ = √(225× 0.36 × 0.64)

⇒ σ = √51.84

⇒ σ = 7.2

Q4: The mean and the standard deviation of a binomial distribution are 100 and 5. Find the binomial distribution.

Solution:

The formula of the standard deviation and mean of binomial distribution is given by:

σ = √(npq), Mean = np

σ = √(Mean × q)

Here, mean = 100 and σ = 5

5 = √(100 × q)

⇒ 25 = 100q

⇒ q = 1/4

Now, p = 1 – q = 1 – 1/4 = 3/4

By mean formula

n = 400 / 3

⇒ n = 133 (approx.)

The binomial distribution is:

P (X = x) = ¹³³C_x (3/4)^x(1/4)^133-x

Q5: Find the mean if the number of good pens is 20 and probability of a good pen is 0.8.

Solution:

The formula of mean in binomial distribution is given by:

Mean = np

Here, n = 20 and p = 0.8

⇒ Mean = 20 × 0.8

⇒ Mean = 16

Q6: If a coin is tossed 4 times. Find the probability that tail appears an odd numbers of times.

Solution:

X be the random variable that tails appear.

The formula to find the above probability is given by:

P (X = x) = ⁿC_x p^x q^n-x

Here,

n = 4, p = (1/2), q = 1 – p = 1/2, x = 1, 3 (odd times)

⇒ P (X = odd) = P(X = 1) + P(X = 3)

⇒ P (X = odd) = ³C₁ (1/2)¹ (1/2)^{3 – 1} + ³C₃ (1/2)³ (1/2)^{3 – 3}

⇒ P (X = odd) = 3(1/2)³ + (1/2)³

⇒ P (X = odd) = 4 / 8

⇒ P (X = odd) = 1/2

Q7: The probability of a man hitting of target is 1/4. If he fires 3 times, what is he probability of his hitting the target at least once.

Solution:

X be the random variable for man hitting target.

The formula to find the above probability is given by:

P (X = x) = ⁿC_x p^x q^n-x

Here,

n = 3, p = (1/4), q = 1 – p = 3/4, x = 1, 2, 3 (hitting at least once)

Now, P (X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3)

⇒ P (X ≥ 1) = ³C₁ (1/4)¹ (3/4)^{3 – 1} + ³C₂ (1/4)² (3/4)^{3 – 2} + ³C₃ (1/4)³ (3/4)^{3 – 3}

⇒ P (X ≥ 1) = 3(1/4) (9/16) + 3(1/16) (3/4) + (1/4)³ (3/4)⁰

⇒ P (X ≥ 1) = 27/64 + 9/64 + 1/64

⇒ P (X ≥ 1) = 37/64

⇒ P (X ≥ 1) = 0.578

Q8: The probability that a student entering a university will graduate is 0.4. Find the probability that out of 3 students at the university none will graduate.

Solution:

X be the random variable for student will graduate.

The formula to find the above probability is given by:

P (X = x) = ⁿC_x p^x q^n-x

Here,

n = 3, p = 0.4, q = 1 – p = 0.6, x = 0 (none of student will graduate)

⇒ P (X = 0) = ³C₀ (0.4)⁰ (0.6)^{3 – 0}

⇒ P(X = 0) = (0.6)³

⇒ P(X = 0) = 0.216

Q9: The mean and variance of a binomial distribution are 8 and 2 then find P (X ≥ 1).

Solution:

The formula for the mean and variance in the binomial distribution is given by:

Mean = np, Variance = npq

Variance = Mean × q

⇒ 2 = 8q

⇒ q = 2 / 8 = 0.25

Now, p = 1 – 0.25 = 0.75

n = Mean /p

⇒ n = 8 / 0.75

⇒ n = 10 (approx.)

Thus, P (X ≤ 1) = P(X = 0) + P(X = 1)

⇒ P (X ≤ 1) = ¹⁰C₀ (0.75)⁰ (0.25)¹⁰ + ¹⁰C₁ (0.75)¹ (0.25)⁹

⇒ P (X ≤ 1) = (0.25)¹⁰ + 9(0.75)¹ (0.25)⁹

⇒ P (X ≤ 1) = (0.25)⁹ [(0.25) + 9(0.75)]

⇒ P (X ≤ 1) = 7 × (0.25)⁹

Q10: Find the variance of the binomial distribution if the mean is 40 and probability of failure is 0.1.

Solution:

The formula to find variance of binomial distribution is given by:

Variance = npq = Mean × q

⇒ Variance = 40 × 0.1

⇒ Variance = 4

Binomial Distribution is a probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials, where each trial has only two possible outcomes: success or failure.

Imagine you’re flipping a coin, but not just once – you’re flipping it many times. Each time, you’re either getting heads or tails. The binomial distribution helps us figure out the chances of getting a certain number of heads (or tails) after flipping the coin a bunch of times.

Here are a few examples of situations that can be modelled using the binomial distribution:

• Suppose you flip a fair coin 10 times. Each flip is an independent trial, and there are only two possible outcomes: heads or tails.
• In a clinical trial, patients are often given a treatment or a placebo. The outcome for each patient might be success (the treatment works) or failure (the treatment doesn’t work).
• A factory produces a large number of items, and each item may be defective or non-defective. Inspectors randomly select a sample of items and check them for defects.
• In an election where voters can choose between two candidates, each voter’s decision can be seen as a trial with two possible outcomes: voting for Candidate A or voting for Candidate B.