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I was doing some questions from my textbook and came across this problem that I'm stuck on how to solve:

Skiers arrive to a top of slope one at a time using a ski lift. The number of arrivals follows a Poisson process with rate λ = 20 arrivals per hour. Suppose that the probability that a skier falls during getting off a ski lift is p = 0.30.

(a)Find the probability that 10 or more skiiers fell in a period of one hour.

(b)Find the probability that no one fell in the first 30 minutes of a ski day.

For (a) I had:

$P(N_f\ge10)$ (where $N_f$ is the Poisson process for no. of falls, which occur at a rate of $20 * 0.3 = 6$ per hour).

$$P(N_f\ge10) = 1-[P(N(1)=0)+P(N(1)=1)+...+P(N(1)=9)]$$ $$= 1-[\frac{(6*1)^0exp(-6)}{0!}+\frac{(6)^1exp(-6)}{1!}+...+\frac{(6)^9exp(-6)}{9!}]$$ Would this way of solving be correct? I feel like there would be a more efficient way to do this. And would I use the same method for (b)?

Thanks in advance!

Raaa
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    for an exact answer, it seems to be the only way to go about it. You can factor the exp(−6) outside the brackets - after which it isn't too bad is it? – Cato Jun 05 '18 at 14:10
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    b) if the process is poisson with $\lambda$ = 6 - then for half hourly periods it is poisson with $\lambda$ = 6 * $\frac{1}{2}$ – Cato Jun 05 '18 at 14:15
  • Yeah that makes sense, I didn't think to factor the exp(-6), makes it easier. Thanks! – Raaa Jun 05 '18 at 14:34

1 Answers1

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For part (a) you are correct.

For part (b) however you must use the exponential distribution.

The poisson process models count (discrete) data and so when asked to consider the waiting time between events (a continuous variable) you must switch to the exponential distribution.

However, when modelling waiting time between poisson modelled events the $\lambda$ parameter for the exponential distribution remains the same as the $\lambda$ parameter for the poisson process in question.

In this case: Find the probability that no one fell in the first 30 minutes of a ski day.

Consider the exponential cdf with parameter $\lambda = 3$ (since it is now 30).

$P(X<x)=1-e^{-\lambda x}$

$P(X>30)=1-P(X<30)$

$P(X>30)=1-[1-e^{-3(0.5)}]$

$P(X>30)=e^{-3(0.5)}$

rmd0001
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