01 Key Concepts
Binomial Conditions
There must be: a fixed number of trials (n), each trial independent, only two outcomes per trial (success/failure), and a constant probability of success (p) each time.
Binomial Formula
P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where C(n,k) is the number of ways to choose k successes from n trials.
Combinations, C(n,k)
C(n,k) = n! / [k!(n-k)!] — counts how many different orders give exactly k successes.
Mean and Variance
For a binomial distribution, mean = n*p, and variance = n*p*(1-p).
When to Use It
Use the binomial distribution whenever you're counting successes across repeated, independent yes/no trials, like defective items or correct guesses.
02 Key Formulas
- P(X=k) = C(n,k) * p^k * (1-p)^(n-k)
- C(n,k) = n! / [k!(n-k)!]
- Mean = n*p
- Variance = n*p*(1-p)
03 Solved Examples
Example 1 A fair coin is flipped 4 times. Find the probability of getting exactly 2 heads.
- Here n=4, k=2, p=0.5.
- C(4,2) = 4!/(2!2!) = 6.
- P(X=2) = 6 * (0.5)^2 * (0.5)^2 = 6 * 0.25 * 0.25.
Answer: P(X=2) = 0.375
Example 2 A test has 5 true/false questions; a student guesses randomly. Find the probability of getting exactly 3 correct.
- Here n=5, k=3, p=0.5.
- C(5,3) = 5!/(3!2!) = 10.
- P(X=3) = 10 * (0.5)^3 * (0.5)^2 = 10 * 0.125 * 0.25.
Answer: P(X=3) = 0.3125
Example 3 For n=10 trials with p=0.3, find the mean and variance of the binomial distribution.
- Mean = n*p = 10*0.3.
- Variance = n*p*(1-p) = 10*0.3*0.7.
Answer: Mean = 3, Variance = 2.1
04 Practice Questions
1A coin is flipped 3 times. Find P(exactly 2 heads).
C(3,2)*(0.5)^2*(0.5)^1 = 3*0.125 = 0.375
2For n=6, p=0.5, find the mean.
3
3For n=8, p=0.25, find the variance.
1.5
4What condition must hold for trials to use the binomial distribution?
Fixed number of independent trials, two outcomes each, constant probability of success
5Find C(5,2).
10
📄 Binomial Distribution — Downloadable Worksheet
10 questions with a full answer key. Grab the PDF to print, or try the interactive version in your browser.