# Random Variable - Poisson + Conditional Probability

I'm new to these topics and struggling to understand the order of attack on this one. I am aware of the formulae for Poisson distribution and Conditional Probability.

The number of times that a person contracts a cold in a given year is a Poisson random variable with parameter $\lambda = 5$. Suppose that a new wonder drug (based on large quantities of vitamin C) has just been marketed that reduces the Poisson parameter to $\lambda = 3$ for $75$ percent of the population. For the other $25$ percent of the population, the drug has no appreciable effect on colds. If an individual tries the drug for a year and has $2$ colds in that time, how likely is it that the drug is beneficial for him or her?

Let $R$ be the event the person is responsive to the drug, and let $T$ be the event $2$ colds. It looks as if we are asked to find $\Pr(R\mid T)$, which is $\frac{\Pr(R\cap T)}{\Pr(T)}$.

We first calculate $\Pr(T)$. With probability $0.75$, the person is in the responds to the drug class, and with probability $0.25$ she is not. Thus $$\Pr(T)=(0.75) e^{-3}\frac{3^2}{2!}+ (0.25)e^{-5}\frac{5^2}{2!}.$$

We leave finding $\Pr(R\cap T)$ to you. Hint: it has already been written down.

More detail: The most crucial thing in this kind of problem is to first identify what probability (conditional or not) we are asked to find. In this case we are asked the probability the treatment is beneficial, *given * the person had two colds. Now we identify the events "two colds" and "beneficial" and assign letters to them. Once we know we want $\Pr(R\mid T)$, we can let the definition of conditional probability guide us.

There are two ways a random person who took the treatment could have two colds. Maybe (i) she responded to the treatment, in other words was in a Poisson with mean $3$ or (ii) she was of the kind unresponsive to the treatment, but despite being a Poisson mean $5$ kind of person, she got lucky.

For $\Pr(T)$, we compute the sum of the probability of (i) and (ii). The probability the person is of the responsive kind is $0.75$. Given that she is of the responsive kind, the probability of two colds is $e^{-3}\frac{3^2}{2!}$. For the probability of (i) we multiply.

• Thanks so much, this was very helpful for me to understand how to approach this problem. I have a follow up question: considering that this is a Poisson distribution, is it already then assumed that all the events are independent, and thus calculating conditional probability is redundant? – Victor Vulovic Jul 30 '16 at 1:36
• The independence associated with a Poisson process is not of importance here. Anyway, in doing the calculation of $\Pr(T)$, we are conditioning on whether the randomly chosen person is of the responsive type or not. I did not write out the reasoning for the calculation of $\Pr(T)$, hoping that the shape of the calculation in the displayed formula would be enough. If you need more detail, I can give it. – André Nicolas Jul 30 '16 at 1:41
• yes, that would be helpful, please. – Victor Vulovic Jul 30 '16 at 1:50

Let $X=$ number of colds. \begin{align} & \Big(\Pr(\text{responsive}\mid X=2),\ \Pr(\text{not responsive} \mid X=2) \Big) \\ = {} & \text{constant} \cdot \Big( \Pr(\text{responsive}),\ \Pr(\text{not responsive}) \Big) \cdot \Big( \Pr(X=2\mid \text{resp.)},\ \Pr(X=2\mid\text{not}) \Big) \end{align} where the two pairs are multiplied componentwise and "constant" means only that it's the same quantity in both compenents of the pair. \begin{align} = {} & c\cdot\left( 3,\ 1 \right) \cdot \left( \frac{e^{-3} 3^2 }{2!},\ \frac{e^{-5} 5^2}{2!} \right) = c\cdot(3,1)\cdot(9e^2,\ 25) = c\cdot(27e^2, 25) \\[10pt] = {} & \left( \frac{27e^2}{27e^2+25},\ \frac{25}{27e^2+25} \right) \approx (0.88864,\ 0.11136). \end{align}