-1
$\begingroup$

Question:

A store opens at $t =0$ and potential customers arrive in a Poisson manner at an average arrival rate of $λ$ potential customers per hour. As long as the store is open, and independently of all other events, each particular potential customer becomes an actual customer with probability $p$. The store closes as soon as ten actual customers have arrived.

Considering only customers arriving between $t =0$ and the closing of the store, what is the probability that no two actual customers arrive within $τ$ time units of each other?

Thanks! Could you give me idea or answer? Why downvote? The given answer is a little werid that I could not understand...

$\endgroup$

closed as off-topic by Did, choco_addicted, user91500, JMP, Watson May 8 '16 at 11:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, choco_addicted, user91500, JMP, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

0
$\begingroup$

I know !!

The probability of no two actual customers arriving within $τ$ time units of each other is equivalent to the probability of all nine independent interarrival times, seperating the ten actual customers, being at least $τ$ time units apart. Thus,

$P$(no two actualcustomers arriving within $τ$ time units of each other) = $$P([T1 ≥ τ]^9)= e^{-9pλτ}$$ .

At first I made a mistake that I just calculated only one interarrival time. However, the whole time is $P(T_{one time})^9$!!

$\endgroup$

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