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I am not sure about my answer. In particular, part b of the following question.

Pizza orders arrive according to a Poisson process of rate 20 per hour. Orders are independently for a vegetarian pizza with probability 1/ 4 , and for a meat pizza with probability 3/4

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a. Six orders arrived between 6:45pm and 7:00pm. Given this, what is the probability that fourteen orders arrive between 7:00pm and 7:45pm?

My Attempt: Let $t=$ time in min. So $\lambda$ per mininute=$20/60=1/3$

$P(N(60)-N(15)=14|N(15)=6)=P(N(45)=14)=\frac{(1/3*45)^{14}}{14!} exp(-1/3*45)$


b. During a particular 60 minute period, 4 vegetarian orders were received. What is the probability that all 4 of them came during the first 30 minutes?

My Attempt:

The conditional poisson would be uniform distributed between (0,60).

Therefore, $P(orders\leq30min)=(30/60)^4=1/16$

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    $\begingroup$ Looks right. Alternately you could compute the conditional probability from first principles. In the first problem, there was no need to convert to minutes, we are dealing with $\frac{3}{4}$ of an hour. $\endgroup$ – André Nicolas Apr 11 '16 at 6:12

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