Binomial Distribution Practice Problems (original) (raw)
Last Updated : 14 Oct, 2025
**Binomial Distribution is a fundamental concept in probability theory , It is a probability distribution that describes the number of successes in a fixed number of independent trials, where each trial has only two possible outcomes: success or failure.
Binomial distribution is widely used across various fields, including statistics, economics, and biology, to model phenomena ranging from coin flips to the success rates of medical treatments, and solving Binomial Distribution Practice Problems is an effective way to understand the application of Binomial Distribution in various scenarios.
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.
Important Formulas on Binomial Distribution
The table below represents the important formulas required to solve the given Binomial distribution Practice Problems.
| Formula Name | Formula |
|---|---|
| Probability Mass Function (PMF) | P (X = x) = **n C x p x q n-x |
| Mean | μ = np |
| Variance | Var(X) = npq |
| Standard Deviation | σ = √(npq) |
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
Binomial Distribution Practice problems - Solved
These binomial distribution practice problems offers significant benefits for understanding and applying this fundamental concept in probability theory.
1: If a coin is tossed 3 times, then find the probability of getting exactly two heads.
X be the random variable for number of heads
The formula for the required probability is given by:
**P (X = x) = **n C 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
2: 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.
The formula for the above probability is given by:
**P (X = x) = **n C 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
3: If the number of trials of a certain binomial distribution is 225 and the probability of success is 0.36. Find the standard deviation.
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
4: The mean and the standard deviation of a binomial distribution are 100 and 5. Find the binomial distribution.
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) = **133 C x (3/4) x (1/4) 133-x
5: Find the mean if the number of good pens is 20 and probability of a good pen is 0.8.
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
6: If a coin is tossed 4 times. Find the probability that tail appears an odd numbers of times.
X be the random variable that tails appear.
The formula to find the above probability is given by:
**P (X = x) = **n C 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
7: 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.
X be the random variable for man hitting target.
The formula to find the above probability is given by:
**P (X = x) = **n C 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
8: 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.
X be the random variable for student will graduate.
The formula to find the above probability is given by:
**P (X = x) = **n C 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
9: The mean and variance of a binomial distribution are 8 and 2 then find P (X ≥ 1).
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) = 1 − P (X=0)
⇒ P (X ≥ 1) = 1 - 10C0 (0.75)0 (0.25)10
⇒ P (X ≥ 1) = 1 − (0.25)10
⇒ P (X ≥ 1) = 1−9.5367×10 -7
⇒ P (X ≥ 1) ≈ 0.999999
**10: Find the variance of the binomial distribution if the mean is 40 and probability of failure is 0.1.
The formula to find variance of binomial distribution is given by:
Variance = npq = Mean × q
⇒ Variance = 40 × 0.1
⇒ Variance = 4