Probability · step 6 of 13

Binomial & uniform distributions

The binomial distribution counts successes in n independent trials each with probability p: P(X = k) = C(n,k)p^k(1−p)^(n−k), with mean np and variance np(1−p). Its four conditions — fixed n, two outcomes, constant p, independence — must all hold, and sampling without replacement breaks the last two.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Check the four conditionsFixed number of trials, binary outcome, constant success probability, independent trials. If any fails, binomial is the wrong model.
Identify n, p and kRead them from the wording. "At least" and "at most" change which k values you sum.
Apply the formula or the complementP(at least one) = 1 − P(0) is far quicker than summing every term.
Use the mean and variance formulasE[X] = np and Var(X) = np(1−p). For the discrete uniform on 1..n, E = (n+1)/2 and Var = (n²−1)/12.

Worked example

A fair coin is tossed 5 times. Find P(exactly 3 heads) and P(at least 1 head).

n = 5, p = 1/2. P(X = 3) = C(5,3)(1/2)³(1/2)² = 10 · (1/32).
= 10/32 = 5/16.
For at least one head, use the complement: P(X = 0) = (1/2)⁵ = 1/32.
P(X ≥ 1) = 1 − 1/32.

Answer. P(exactly 3) = 5/16 and P(at least 1) = 31/32.

Where marks get dropped

These are the specific errors that cost credit on binomial & uniform distributions questions — QED's rubric penalises each of them separately.

Using the binomial for sampling without replacement. That is hypergeometric — p changes between draws.
Omitting the binomial coefficient C(n,k), which counts the orderings of the successes.
Misreading "at least 3" as "exactly 3". At least 3 requires summing k = 3, 4 and 5.

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Binomial & uniform distributions — frequently asked questions

When is a binomial approximately normal?

When np and n(1−p) both exceed about 10. Then X ≈ N(np, np(1−p)), which is the basis of the standard proportion tests.

What if trials are not independent?

Use the hypergeometric distribution for sampling without replacement from a finite population, or model the dependence explicitly.

What is the discrete uniform distribution?

Every value equally likely, like a fair die. On 1..n the mean is (n+1)/2 and the variance is (n²−1)/12, which gives 35/12 for n = 6.

The rest of Probability

Events, conditional probability, random variables. Each subtopic below has its own method, worked example and mark-losing traps.

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