I need some help with this one. "The numbers of calls in a call center has the Poisson distribution. It is known that in a 6 days week the average number of calls is 12. What is the probability that on Tuesday there will be at least one call, given that in exactly 2 days during this week there was no call at all ? " I tried defining a Binomial random variable of number of days with no call at all (after calculating the probability of this event using Poisson distribution). I managed to find what I believe is the denominator of the solution, but I am stuck with the nominator. Can you please assist me with this problem ? Thank you !
[Math] Number of calls in a call center
binomial distributionpoisson distributionprobability
Best Answer
Grinding it out: Let $A$ be the event there was at least one call on Tuesday, and $B$ be the event there were exactly two days with no call. We want $\Pr(A\mid B)$. It is easier to think about $\Pr(A^c\mid B)$.
Let $p$ be the probability of no call on a particular day. You know how to find $p$.
Then the probability of exactly two call-free days in a "week" is $\binom{6}{2}p^2(1-p)^4$.
To finish calculating $\Pr(A^c\mid B)$, we need $\Pr(A^c\cap B)$. This is the probability of no call on Tuesday, times the probability exactly one of the other $5$ days is call-free. This is $p\binom{5}{1}p(1-p)^4$.
Divide. There is cancellation and we get $\frac{2}{6}$.
Another way: If there were two call-free days, the probability one of them was Tuesday is obviously $\frac{2}{6}$. So the conditional probability of at least one call on Tuesday is $\frac{4}{6}$.