# Calculating probability from Poisson random variable

The number of times an individual contracts a cold in a given year is a Poisson random variable with parameter $\lambda = 2$. Suppose that a new wonder drug has just been marketed that reduces the Poisson parameter to $\lambda = 1$ for 80% of the population. For the other 20% of the population the drug has no appreciable effects on cold. If an individual tries the drug for a year and has 0 colds in that time, how likely is that the drug is beneficial for him/her?

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Let $B$ be the event "the drug is beneficial" (to our randomly selected individual), and let $N$ be the event "$0$ colds." We want to find $\Pr(B|N)$. By the usual formula, we have $$\Pr(B|N)=\frac{\Pr(B\cap N)}{\Pr(N)}.$$ We need to evaluate the probabilities on the right.

The individual can have $0$ colds in the year in two disjoint ways: if (i) she is in the $80\%$ of the population that benefits, and she gets $0$ colds or (ii) she is in the $20\%$ that does not benefit, and yet get $0$ colds.

For (i), with probability $0.8$ she benefits. Given that she benefits, the probability of $0$ colds is $e^{-1}$. So the probability of (i) is $(0.8)e^{-1}$.

For (ii), with probability $0.2$ she does not benefit. In that case, the probability of $0$ colds is $e^{-2}$. So the probability of (ii) is $(0.2)e^{-2}$.

Thus $$\Pr(N)=(0.8)e^{-1}+(0.2)(e^{-2}.$$

We already calculated $\Pr(B\cap N)$. It is just $(0.8)e^{-1}$, the probability of (i).

Finally, divide. We get $$\Pr(B|N)=\frac{(0.8)e^{-1}}{(0.8)e^{-1}+(0.2)e^{-2}}.$$

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