The probability that a store will have exactly k customers on any given day is $$P_K(k)=\frac{1}{5}\left(\frac{4}{5}\right)^k, \quad k=0,1,2,3....$$ Every day, out of all the customers who purchased something from the store that day, one is randomly chosen to win a prize. Every customer that day has an equal chance of being chosen. Assume that no customer visits this store more than once a day, and further assume that the store can handle an infinite number of customers.

  • (c). Given a customer who has won a prize, what is the probability that he was in the store on a day when it had a total of exactly $k$ customers?

  • (d).Since the store owner’s birthday is in July, he decides to celebrate by giving two prizes each day in the month of July. At the end of each day, after one customer is randomly chosen to win the first prize, another winner is randomly chosen from the remaining $k − 1$ customers, and given the second prize. No one can win both prizes. Let $X$ denote the customer number of the first winner, and let $Y$ denote the customer number of the second winner. Determine the joint pdf, $P(x,y|k), for 1 ≤ x, y ≤ k$.

-- My answer for (c) is $P=\frac{1}{5}(\frac{4}{5})^k\frac{1}{k}$,not sure if it is right. For the part (d),don't really know how to do that.


Let's call the event of a price $pr$.

You're right that $P(pr \land K=k) = \frac{1}{k}\frac{1}{5}(\frac{4}{5})^k$ for $k > 0$: first we need we need $k$ customers and then there is a $\frac{1}{k}$ chance that this customer won it. Also $P(pr | K=k) = \frac{1}{k}$.

But the asked for probability is $P(X=k|pr)$ so this reeks of Bayes' law, which comes down to (in this case, where the different $K=i$ are the mutually disjoint events):

$$P(K=k|pr) = \frac{P(pr|K=k)P(K=k)}{\sum_{i=0}^\infty P(pr|K=i)P(K=i)}$$

and the numerator is your $\frac{1}{k}\frac{1}{5}(\frac{4}{5})^k$

and the denominator is (the term for $i=0$ equals $0$, no customers, no price) $$ \sum_{i=1}^{\infty} \frac{1}{i} \frac{1}{5}(\frac{4}{5})^i$$.

Now it's just calculus. Remember the series $$\sum_{n\ge 1} \frac{1}{n} x^n = \sum_{n \ge 0} \left(\int (x^{n} dx\right) =\int \left(\sum_{n \ge 0} x^n\right) dx$$ etc.

As to (d), this seems simpler. $P(X=x, Y=y | K=k)$ is only non-zero for $k \ge 2$ and $x \neq y$ and $1 \le x,y \le k$ and it that case equals $\frac{1}{k}\frac{1}{k-1}$ (there is a $1$ in $k$ chance that $x$ will be chosen as winner $1$, and then a $1$ over $k-1$ chance that $y \neq x$ will be chosen as winner nr $2$. Note that given $k$ there are $k(k-1)$ possible winner pairs all of which have the same chance.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.