= Solution
The <law of total probability> and the definition of <conditional probability> give <Bayes' theorem>:
$$
P(A_i\mid E)=\frac{P(A_i\cap E)}{P(E)}
=\frac{P(A_i)P(E\mid A_i)}{\sum_{j=1}^3P(A_j)P(E\mid A_j)}.
$$
The denominator is positive because $P(E)>0$.
For the observed cargo, set $p_A=0.05$, $p_B=0.03$, $p_C=0.01$. The <binomial likelihood> under origin $i$ is
$$
L_i=P(E\mid A_i)=\binom{10000}{200}p_i^{200}(1-p_i)^{9800}.
$$
Equal prior probabilities and <Bayes' theorem> make the posterior proportional to these likelihoods. The binomial coefficient cancels in the <likelihood ratios>:
$$
\begin{aligned}
\log\frac{L_A}{L_B}
&=200\log\frac53+9800\log\frac{0.95}{0.97}\approx-102.00893,\\
\log\frac{L_C}{L_B}
&=-200\log3+9800\log\frac{0.99}{0.97}\approx-19.71552.
\end{aligned}
$$
Thus $L_A/L_B\approx4.99\times10^{-45}$ and $L_C/L_B\approx2.74\times10^{-9}$, giving
$$
\boxed{P(B\mid E)=\frac1{1+L_A/L_B+L_C/L_B}
\approx0.99999999726.}
$$
Even the permitted approximation $\log(1-p)\approx-p$ gives log ratios about $-93.83$ and $-23.72$, which already show overwhelming posterior preference for B. The exact logarithms are preferable because multiplication by $9800$ magnifies the approximation error. An observed contamination fraction halfway between two hypothesized rates does not give equal <likelihoods>: the <binomial distribution> assigns asymmetric probabilities to those deviations, especially at this large sample size.
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