The number of emails that I get in a weekday (Monday through Friday) can be modeled by a Poisson distribution with an average of 1/6 emails per minute. The number of emails that I receive on weekends (Saturday and Sunday) can be modeled by a Poisson distribution with an average of 1/30 emails per minute
A random day is chosen (all days of the week are equally likely to be selected), and a random interval of length one hour is selected in the chosen day. It is observed that I did not receive any emails in that interval. What is the probability that the chosen day is a weekday?
Best Answer
Let $X$ represent the number of emails you receive in an hour on a weekday and let $Y$ represent the number of emails you receive in an hour on a weekend. Let $Z$ be the random variable representing how many emails you received in an hour knowing that the day you selected could be any day of the week.
$$X\sim\mathrm{Po}(10)\implies\operatorname{P}(X=x)=\frac{10^x}{e^{10}x!}\tag{1}$$ $$Y\sim\mathrm{Po}(2)\implies\operatorname{P}(Y=y)=\frac{2^y}{e^{2}y!}\tag{2}$$
Bayes’ theorem tells us
$$\operatorname{P}(B \,|\, A) = \frac{ \operatorname{P}(B) \, \operatorname{P}(A \,|\, B) }{\operatorname{P}(B) \, \operatorname{P}(A \,|\, B) + \operatorname{P}\left(B'\right) \, \operatorname{P}\left(A \,\middle|\, B'\right)}$$
Taking $B$ to be the event $D$ and $A$ to be the event $Z=0$, we have
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(Z=0 \,|\, D) }{\operatorname{P}(D) \, \operatorname{P}(Z=0 \,|\, D) + \operatorname{P}\left(D'\right) \, \operatorname{P}\left(Z=0 \,\middle|\, D'\right)}$$
Given if $D$ occurs, $Z=0$ can actually be more precisely written as $X=0$; if $D'$ occurs, then $Z=0$ can actually be more precisely written as $Y=0$, giving you
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(X=0 \,|\, D) }{\operatorname{P}(D) \, \operatorname{P}(X=0 \,|\, D) + \operatorname{P}\left(D'\right) \, \operatorname{P}\left(Y=0 \,\middle|\, D'\right)}$$
By the definition of our variables, we can further contract this to
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(X=0) }{\operatorname{P}(D) \, \operatorname{P}(X=0) + \operatorname{P}\left(D'\right) \, \operatorname{P}(Y=0)}$$
because if we’re analyzing $X$, we know we’re looking at a weekday, and if we’re analyzing $Y$, we know we’re looking at a weekend.
Use $(1)$ and $(2)$ in conjunction with $\operatorname{P}(D)=5/7$ and $\operatorname{P}\left(D'\right)=2/7$ to arrive at your answer.