Binomial Probability
Probability of k successes in n independent trials.
Formula
Variables
Example
3 heads in 5 flips: C(5,3)(0.5)⁵ = 0.3125
How it works
This formula gives the probability of getting exactly k successes in n independent trials when each trial has the same success chance p. It works by combining two ideas: there are \u201cchoose\u201d many different ways to place the k successes among the n trials, and any one particular arrangement has probability p^k(1-p)^{n-k} because the trials are independent. Multiplying these together counts every valid arrangement once and only once.
When to use it
Use it when each trial has only two outcomes, the trials do not affect each other, and the probability of success stays constant from trial to trial. Typical examples include coin flips, defective parts from a stable production line, survey responses with a fixed yes-rate, or passing a repeated test with a constant chance each time.
Worked example, step by step
Find the probability of exactly 2 rainy days in a 4-day period if the chance of rain each day is 0.3. P(X=2) = C(4,2)(0.3)^2(1-0.3)^(4-2) P(X=2) = 6(0.3)^2(0.7)^2 P(X=2) = 6(0.09)(0.49) P(X=2) = 6(0.0441) P(X=2) = 0.2646 Final answer: 0.2646 (about 26.46%)
Common mistakes to avoid
- Forgetting the combination factor C(n,k), which counts how many ways the successes can be arranged
- Using the wrong exponent on p or (1-p), especially switching k and n-k
- Applying the formula when the trials are not independent or the success probability changes from trial to trial
Did You Know?
The binomial distribution models repeated yes/no experiments.
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