Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — a single binomial probability
X∼B(10,0.3). Find P(X=3).
- 1P(X=r)=(rn)prqn−r with n=10, r=3, p=0.3, q=0.7.
- 2(310)=120The number of orders in which 3 successes can fall among 10 trials.
- 3(0.3)3=0.027The three successes.
- 4(0.7)7=0.08235The seven failures. The exponents add to 10 ✓
- 5120×0.027×0.08235=0.267 (3 s.f.)Close to the mean of 3, so a relatively large probability.
Example 2 — using the complement
X∼B(10,0.3). Find P(X≥1).
- 1X≥1 covers ten separate values — a long calculation directly.
- 2Its complement is a single term: X=0.This is the shortcut worth looking for every time.
- 3P(X=0)=(010)(0.3)0(0.7)10=(0.7)10Both the combination and the zero power give 1.
- 4=0.0282
- 5P(X≥1)=1−0.0282=0.972Nearly certain — with ten trials at 30%, at least one success is very likely.
Example 3 — setting up a binomial model
A player scores 70% of free throws and takes 12. Find the probability of scoring exactly 9, and state an assumption.
- 1A trial is one free throw; success is scoring.Define the trial and the success explicitly.
- 2n=12 is fixed and p=0.7 is constant.Two of the four conditions.
- 3Assume the throws are independent — the player's form does not change with success or failure.The condition most often questionable in practice.
- 4X∼B(12,0.7)
- 5P(X=9)=(912)(0.7)9(0.3)3=220×0.04035×0.027
- 6=0.240 (3 s.f.)The mean is 12×0.7=8.4, so 9 is a likely outcome ✓