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...


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

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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 $P(T_{one time})^9$!!


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