Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — a geometric probability and mean
15% of tyres are faulty. Tyres are tested one at a time until a faulty one is found. Find P(X=4) and E(X).
- 1X∼Geo(0.15), so p=0.15 and q=0.85Counting trials until the first success identifies the geometric.
- 2P(X=4)=q3p=0.853×0.15Three failures, then the success.
- 3=0.0921
- 4E(X)=p1=0.151=6.67A long-run average, so a decimal is correct.
Example 2 — a Poisson probability over a changed interval
Calls arrive at a switchboard at an average of 3 per five minutes. Find the probability of exactly 10 calls in fifteen minutes.
- 1Fifteen minutes is three intervals, so λ=3×3=9Scale λ before anything else.
- 2P(X=10)=10!e−9910
- 3910=3486784401 and 10!=3628800
- 4P(X=10)=0.1186Close to λ, as expected — the mode of a Poisson sits near its mean.
Example 3 — approximating a binomial by a Poisson
A component has a 1% failure rate and 200 are installed. Use a suitable approximation to find P(X≥2).
- 1n=200>50 and np=2<5, so a Poisson approximation is validState the check — it carries a mark.
- 2X≈Po(2)λ=np.
- 3P(X≥2)=1−P(X=0)−P(X=1)Two terms only — far quicker than summing upwards.
- 4=1−e−2−2e−2=1−3e−2
- 5=1−0.4060=0.5940