# Customers arriving in a store, find $P(X=0).$

Customers arrive to a store such that the number of arriving customers in an hour has a Poisson distribution with mean $4$. A customer is male or female with equal probabilities. Let $X$ be the number of female customers in an hour and find $P(X = 0).$

So I have that $$X=\text{Number of female visitors.} \\Y=\text{Number of total visitors.} \ \ \ \\ Y\sim\text{Poi}(4) \quad \quad \quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$

And I'm looking for the probability that no females visited the store in a given hour.

If $X$ and $Y$ are discrete random variables and have joint pmf $p$, the conditional pmf of $Y$ and given $X=j$ is defined as

$$p_Y(Y=k|X=j)=\frac{p(X=j,Y=k)}{p_X(X=j)}. \tag{1}$$

Rewriting this I get

$$p_X(X=j)=\frac{p(X=j,Y=k)}{p_Y(Y=k|X=j)}.$$

I'm not sure but I believe that $p(x_j,y_k)=e^{-4}4^0/0!=1/e^4.$ To compute $p_X(x_j)$ for $j=0$ I also need $p_Y(y_k|x_j).$ How do I find this out?

Am I even doing this correctly to begin with?

• I think using symmetry between both genders can solve the problem. – GNUSupporter 8964民主女神 地下教會 Apr 24 '18 at 18:26
• If you get an average of $4$ customers per hour, and half of them are female, then you get an average of how many female customers per hour? – G Tony Jacobs Apr 24 '18 at 18:28
• @GTonyJacobs - That would be 2 females/hour. Crap, is it that easy? This was under the chapter conditional distributions and independence. Where do these concepts come in to play in this problem? – Parseval Apr 24 '18 at 18:30
• @Parseval it is but it requires that we be using a Poisson model of arrivals (which of course we are). See my answer for the approach with conditional probability. – spaceisdarkgreen Apr 24 '18 at 18:38

There are two ways to approach this. First, for any number of customer arrivals you can compute the probability that none are women. Then you can use law of total probability to express the probability no women arrive as a sum. You appear to have chosen this approach and then fallen off the rails (seem to have forgotten there is a sum involved.) We have $P(X=0\mid Y=n)=1/2^n,$ so $$P(X=0)= \sum_n P(X=0\mid Y=n)P(Y=n)=\sum_n \frac{1}{2^n} \frac{4^ne^{-4}}{n!}$$
• Correct me if I'm wrong but the thought behind the probability for no women, given $n$ customers is that the probability that the first customer is a women is $1/2,$ second customer $(1/2)^2,...$ $n$th customer is $(1/2)^n$ right? Can you elaborate on the application of the law of total probability? Also, shouldn't the sums go from n=1 to infinity? What does it mean only having an $n$ at the bottom? – Parseval Apr 24 '18 at 18:56
• Just having an $n$ at the bottom (sum over all $n$) is the same as saying "all possible values of $n$", which could be $1$ to infinity, or $0$ to infinity, or whatever makes sense in context. – G Tony Jacobs Apr 24 '18 at 19:19
• @Parseval Yeah, that’s the reasoning for $1/2^n.$ The $n$ ranges over all the possible values for $Y,$ i.e. from $0$ to $\infty.$ The law of total probability says that if events $B_i$ partition all the possible outcomes and $A$ is an event then $P(A)=\sum_i P(A,B_i).$ Here the partition chosen were the different values the total number of arrivals $Y$ could take, i.e. the events $\{Y=n\}$ for $n=0,1,2,\ldots.$ – spaceisdarkgreen Apr 24 '18 at 19:24
• Thanks for the reply! May I ask, what is the intuitive difference between $P(X=0)$ and $P(X=0|Y=n)$? They seem quite the same to me. – Parseval Apr 24 '18 at 19:31