# Paper 2, Section II, F

Let $A_{1}, A_{2}$ and $A_{3}$ be three pairwise disjoint events such that the union $A_{1} \cup A_{2} \cup A_{3}$ is the full event and $\mathbb{P}\left(A_{1}\right), \mathbb{P}\left(A_{2}\right), \mathbb{P}\left(A_{3}\right)>0$. Let $E$ be any event with $\mathbb{P}(E)>0$. Prove the formula

$\mathbb{P}\left(A_{i} \mid E\right)=\frac{\mathbb{P}\left(A_{i}\right) \mathbb{P}\left(E \mid A_{i}\right)}{\sum_{j=1,2,3} \mathbb{P}\left(A_{j}\right) \mathbb{P}\left(E \mid A_{j}\right)}$

A Royal Navy speedboat has intercepted an abandoned cargo of packets of the deadly narcotic spitamin. This sophisticated chemical can be manufactured in only three places in the world: a plant in Authoristan (A), a factory in Bolimbia (B) and the ultramodern laboratory on board of a pirate submarine Crash (C) cruising ocean waters. The investigators wish to determine where this particular cargo comes from, but in the absence of prior knowledge they have to assume that each of the possibilities A, B and C is equally likely.

It is known that a packet from A contains pure spitamin in $95 \%$ of cases and is contaminated in $5 \%$ of cases. For B the corresponding figures are $97 \%$ and $3 \%$, and for $\mathrm{C}$ they are $99 \%$ and $1 \%$.

Analysis of the captured cargo showed that out of 10000 packets checked, 9800 contained the pure drug and the remaining 200 were contaminated. On the basis of this analysis, the Royal Navy captain estimated that $98 \%$ of the packets contain pure spitamin and reported his opinion that with probability roughly $0.5$ the cargo was produced in B and with probability roughly $0.5$ it was produced in C.

Assume that the number of contaminated packets follows the binomial distribution $\operatorname{Bin}(10000, \delta / 100)$ where $\delta$ equals 5 for $A, 3$ for $B$ and 1 for C. Prove that the captain's opinion is wrong: there is an overwhelming chance that the cargo comes from B.

[Hint: Let $E$ be the event that 200 out of 10000 packets are contaminated. Compare the ratios of the conditional probabilities $\mathbb{P}(E \mid A), \mathbb{P}(E \mid B)$ and $\mathbb{P}(E \mid C)$. You may find it helpful that $\ln 3 \approx 1.09861$ and $\ln 5 \approx 1.60944$. You may also take $\ln (1-\delta / 100) \approx-\delta / 100$.]