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bixtya [17]
3 years ago
12

Suppose customer arrivals at a post office are modeled by a Poisson process N with intensity λ > 0. Let T1 be the time of the

first arrival. Let t > 0. Suppose we learn that by time t there has been precisely one arrival, in other words, Nt = 1. What is the distribution of T1 under this new information? In other words, find the conditional probability P(T1 ≤ s|Nt = 1) for all s ≥ 0.
Mathematics
1 answer:
Lynna [10]3 years ago
4 0

Answer:

Step-by-step explanation:

We need to find the conditional probability P( T1 < s|N(t)=1 )  for all s ≥ 0

P( time of the first person's arrival < s till time t exactly 1 person has arrived )

= P( time of the first person's arrival < s, till time t exactly 1 person has arrived ) / P(exactly 1 person has arrived till time t )

{ As till time t, we know that exactly 1 person has arrived, thus relevant values of s : 0 < s < t }

P( time of the first person arrival < s, till time t exactly 1 person has arrived ) / P(exactly 1 person has arrived till time t )

= P( exactly 1 person has arrived till time s )/ P(exactly 1 person has arrived till time t )

P(exactly x person has arrived till time t ) ~ Poisson(kt) where k = lambda

Therefore,

P(exactly 1 person has arrived till time s )/ P(exactly 1 person has arrived till time t )

= [ kse-ks/1! ] / [ kte-kt/1! ]

= (s/t)e-k(s-t)

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Steps:

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Description:

The person will spend less than 20 or o.2 as seen in the question. Meaning p will equal 0.2. P for probability. Now n is for number. n will equal 17. And then we have to times all the number. N times P will equal 17 then times it by 0.2=  Answer will come as 3.4.

Answer: 3.4

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This is further explained below.

<h3>What is mAngleVSR?</h3>

Generally, Draw two lines: one that connects the points R, S, and U, and another that connects the points V, S, and T. (see attached diagram). At point S, these lines come together to create four angles, which are denoted by the letters RSV, VSU, UST, and TSR respectively.

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