Binomial Distribution: From Intuition to Precision
Imagine you flip a fair coin 10 times. You want to know the probability of getting exactly 6 heads. This is the kind of question the binomial distribution answers — it models the number of "successes" in a fixed number of independent trials, where each trial has only two outcomes.
The Intuition
Think of a single trial. You roll a die and call "getting a 4" a success. That's one trial. Now roll the die 5 times. The number of times you get a 4 could be 0, 1, 2, 3, 4, or 5. Each roll is independent — the result of one roll doesn't affect the next. The probability of success (getting a 4) stays the same every time: p=61.
The binomial distribution tells you exactly how likely each possible count of successes is, given:
- a fixed number of trials n,
- a constant success probability p per trial,
- independent trials.
The Key Conditions
For a situation to be modelled by a binomial distribution, all four must hold:
- Fixed number of trials (n). You decide in advance how many times you'll repeat the experiment.
- Two outcomes per trial — "success" and "failure". These are just labels; success is whatever outcome you're counting.
- Constant probability of success (p) on every trial.
- Independent trials. The outcome of one trial does not influence another.
A common mistake is to apply the binomial distribution when trials are not independent — for example, drawing cards without replacement from a deck. That's a hypergeometric situation, not binomial.
The Precise Statement
Let X be the number of successes in n independent trials, each with success probability p. Then X follows a binomial distribution with parameters n and p, written:
X∼Binomial(n,p)
The probability of getting exactly k successes (where k=0,1,2,…,n) is:
P(X=k)=(kn)pk(1−p)n−k
P(X=k)=(kn)pk(1−p)n−k
Breaking Down the Formula
Three pieces multiply together:
- pk — the probability that the k successes happen. Since each success has probability p, and there are k of them, this factor is p multiplied by itself k times.
- (1−p)n−k — the probability that the remaining n−k trials are failures. Each failure has probability 1−p.
- (kn) — the number of ways to choose which k of the n trials are the successes. The successes could be the first k trials, or the last k, or any arrangement. This binomial coefficient counts all possible arrangements.
The binomial coefficient (kn) is read as "n choose k" and equals k!(n−k)!n!. For example, (25)=2!3!5!=10.
A Worked Example
Problem: A multiple-choice test has 10 questions, each with 4 options. You guess every answer. What is the probability you get exactly 3 correct?
Solution:
- n=10 (10 trials)
- p=41 (probability of guessing correctly on one question)
- k=3 (we want exactly 3 successes)
P(X=3)=(310)(41)3(43)7
Compute (310)=120, so:
P(X=3)=120×641×(43)7
(43)7=163842187, so:
P(X=3)=120×641×163842187=64×16384120×2187
Simplify: 120/64=15/8, so:
P(X=3)=8×1638415×2187=13107232805≈0.2503
So there's about a 25% chance of getting exactly 3 correct by pure guessing.
Mean and Variance
For a binomial distribution, two important summary measures are: …