Binomial Distribution Practice Problems

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.

The table below represents the important formulas of Binomial distribution.

Where,

• n is number of trials
• p is probability of success
• q is probability of failure
• μ is mean or expected value
• Var(X) is variance
• σ is standard deviation

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

Q1: If a coin is tossed 5 times find the probability of getting at most 2 heads.

Q2: A pair of dice is thrown 7 times. If getting a total of 11 is considered as a success, what is the probability of getting at least 3 success.

Q3: If mean of the binomial distribution is 20 and the number of observations is 30. Find the variance and standard deviation.

Q4: Find the mean of the binomial distribution given that number of trials is 80 and probability of success is 0.45.

Q5: Calculate the variance of the binomial distribution given that number of trials is 200 and probability of success is 0.8.

Q6: If the mean and variance of binomial distribution are 45 and 30. Find the binomial distribution.

Q7: The mean and variance of a binomial distribution are 6 and 2 then find P(X>3).

Q8: The probability of a boy hitting a target is 0.2. How Many times must he fire so that the probability of his hitting the target at least once is greater than 2/3.

Q9: Find the probability distribution of the number of heads when 5 coins are tossed.

Q10: There are 20 machine and that the chance of any one of them to be out of service is 0.025. Determine the probability that exactly three machines will be out of service on the same day.

What is Binomial Distribution and its Properties?

Binomial distribution is probability distribution that calculates the probability of x success by using number of trials, success probability and failure probability.

What are the 4 Factors of Binomial Distribution?

The 4 factors of binomial distribution are: number of trials, number of successes, probability of success and probability of failure.

What is the Main Formula of Binomial Distribution?

The main formula of Binomial distribution is given by:

P (X = x) = ⁿC_x p^x q^n-x

How to Find q in Binomial Distribution?

To find q i.e., probability of failure in binomial distribution we calculate 1 – p where, p is the probability of success.

What is p in Binomial Distribution?

p in Binomial distribution is the probability of success.