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A salesperson visits $k$ clients each day. The salesperson makes one and exactly one sale each day. The probability that the salesperson makes a sale to the $j$th client on a given day is $p_j$, independent of previous days. Since the salesperson makes one and exactly one sale each day, $p_1 + p_2 + … +p_k = 1$. What is the probability that sale made on the $N$th day is to client that is different than on any of the previous $N-1$ days?

A hint given at the end of this question was to use Conditional Probability to solve this question.

My attempt:

Suppose salesperson selects $i$th client on $N$th day. We need to calculate the probability that $i$th client is not selected in any of the $N-1$ days before. Probability of an $i$th client not being selected on a day is given by $1-p_i$. Hence, for $N-1$ days it will be $(1-p_i)^{N - 1}$. So the net probability to select $i$th client on $N$th day would be $p_i(1-p_i)^{N - 1}$.

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    $\begingroup$ What is your attempt on the problem and where are you having difficulties? $\endgroup$
    – anak
    Commented Jul 10, 2015 at 15:36
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    $\begingroup$ Thanks for the reply. My analysis is as follow.. On the Nth day, the sales person can select ith client with probability pi. But the condition is that ith client should not have been picked in any of the earlier N-1 days. probability for this event would be (1-pi) raised to (N-1). Hence the net probability would pi. (1 - pi) raised to (N-1). But I am not convinced that this can be so simple. $\endgroup$
    – naresh
    Commented Jul 10, 2015 at 15:37
  • $\begingroup$ Do you mean that the salesman sees the same k clients every day? Assuming this, here is a hint: first answer the simpler question "what is the probability that client #1 is chosen on day N for the first time?" $\endgroup$
    – lulu
    Commented Jul 10, 2015 at 16:40
  • $\begingroup$ The salesman sees the same k clients every day. For client#1 to be chosen on Nth day for the first time, he should not have been chosen on the N-1 days before. If p1 is the probability with which client#1 gets chosen, will this not be (1-p1)raised to (N-1)?? $\endgroup$
    – naresh
    Commented Jul 10, 2015 at 16:55
  • $\begingroup$ Pls help me in understanding this question. Thanks $\endgroup$
    – naresh
    Commented Jul 10, 2015 at 19:51

1 Answer 1

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Your calculation is correct; now you just have to sum that over $i$.

It's interesting, by the way, that this calculation is valid despite the correlation introduced by the condition that exactly one client is selected every day. The probabilities are independent across days but not across clients. You might want to think about that a bit and include in your answer an explanation why this doesn't invalidate your approach.

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