- #1

Mehmood_Yasir

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## Homework Statement

Pedestrians approach to a signal at the crossing in a Poisson manner with arrival rate ##\lambda## arrivals per minute. The first pedestrian arriving the signal starts a timer ##T## then waits for time ##T##. A light is flashed after time T, and all waiting pedestrians who arrive within time duration ##T## must cross.

What is the probability that a randomly arriving padestrian has crossed the crossing in a group of ##k+1## padestrians? Here is the group is called ##k+1## because first one starts ##T## and ##k## more padestrian arrive within time ##T##.

The answer given is ##\frac{(K+1) {(\lambda * T)}^k e^{-\lambda T} } {k! (1+\lambda T)}##

I could not understand this answer. Can someone kindly explain me this answer.

## Homework Equations

Poisson formula for general ##k## arrivals in time ##T##,

##P_k= \frac{{(\lambda * T)}^k e^{-\lambda T} } {k! }##

## The Attempt at a Solution

Since the total padestrian which has crossed the crossing after light is flashed = ##k+1##

We know that for a Poisson process, the probability of ##k## arrivals in a given time interval ##T## is ## \frac{{(\lambda * T)}^k e^{-\lambda T} } {k! }##.

Probability that ##k+1## has crossed the crossing is ## \frac{{(\lambda T)}^{k+1} e^{-\lambda T} } {(k+1)! }##

What is the probability that a randomly arriving padestrian has crossed the crossing in a group of ##(k+1)## ??