Find the probability that any particular 200 unit batch violates the guarantee
- exactly, using a binomial distribution
- using the Poisson approximation
2026-04-05 17:18:34.1775409514
A distributor of TVs determines that 1% of them are faulty. He sell them in bulk of 200 units and guarantees that 98% of the TVs won't be faulty.
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Sketch The number of faulty TVs is $X\sim B(200,\,0.01)\approx\operatorname{Poisson}(2)$. The guarantee is honoured iff $X\le4$, with probability $\sum_{k=0}^4\binom{200}{k}0.01^k0.99^{200-k}\approx e^{-2}\sum_{k=0}^4\frac{2^k}{k!}$. Once either expression is calculated by summing just $5$ terms, subtraction from $1$ gives the probability of guarantee violation. By my calculation, the exact result $0.05175$ admits Poisson approximation $0.05265$.