What is the binomial distribution (in one sentence)?
The binomial distribution gives the probabilities of getting exactly k "successes" in n independent trials when each trial has only two outcomes (success/failure) and the probability of success is the same in every trial.
Key conditions (when you can use it)
- Fixed number of trials: n is set in advance.
- Only two outcomes per trial: call them success and failure.
- Each trial has the same probability of success: p (and failure probability 1 - p).
- Trials are independent: the outcome of one trial doesn't change the others.
Probability formula (the core idea)
Let X be the number of successes in n trials. For k = 0, 1, 2, ..., n:
P(X = k) = C(n, k) p^k (1 - p)^(n - k)
Explanation of each part:
- p^k: the probability that a particular set of k specific trials are successes.
- (1 - p)^(n - k): the probability the remaining n - k trials are failures.
- C(n, k) = n! / (k! (n - k)!): the number of different ways to choose which k trials (out of n) are the successes. We multiply by this because the successes could occur in any of those combinations.
Simple example — coin flips
Example: Flip a fair coin (p = 0.5 for heads) 10 times (n = 10). What is the probability of exactly 4 heads (k = 4)?
- Compute the combination: C(10, 4) = 210.
- Compute p^k (1 - p)^(n - k) = 0.5^4 * 0.5^6 = 0.5^10 = 1/1024.
- Multiply: P(X = 4) = 210 * 1/1024 ≈ 0.2051 (about 20.5%).
Intuition
If you want k successes, you need any specific pattern of k successes and n-k failures (probability p^k (1-p)^(n-k)). But there are many patterns (C(n,k)), so add them up by multiplying. That gives the full probability of 'exactly k successes'.
Mean and variance (what to expect)
- Expected number of successes: E[X] = n p.
- Variance: Var(X) = n p (1 - p).
Why? Think of X as the sum of n independent Bernoulli trials (each 1 for success, 0 for failure). Each trial has mean p and variance p(1-p). Summing n such independent trials gives mean np and variance n p(1-p).
Other useful points
- To compute "at least one" success: P(X ≥ 1) = 1 - P(X = 0) = 1 - (1 - p)^n.
- If n is large and p is not too close to 0 or 1, the binomial can be approximated by a normal distribution with mean np and variance np(1-p) (use continuity correction when needed).
- If n is large and p is small with λ = n p moderate, a Poisson(λ) approximation can be used: P(X = k) ≈ e^{-λ} λ^k / k!.
- Do not use the binomial if trials are not independent, probabilities change between trials, or outcomes are not just success/failure.
Quick checklist to recognize a binomial problem
- Is there a fixed number n of trials?
- Are outcomes only success/failure per trial?
- Is the success probability the same on every trial (p)?
- Are trials independent?
If the answer is yes to all four, use the binomial formula.
If you want, give me a concrete example (dice, tests, coin flips, surveys) and I will compute probabilities and walk through the steps with numbers.