A Poisson Question

Question

IB Poisson Question

A ferry carries cars across a river. There is a fixed time of $T$ minutes between crossings. The arrival of cars at the crossing can be assumed to follow a Poisson distribution with a mean of one car every four minutes. Let $X$ denote the number of cars that arrive in $T$ minutes.

(a) Find $T$ , to the nearest minute, if $P(X ≤ 3) = 0.6$. [3 marks]

It is now decided that the time between crossings, $T$ , will be $10$ minutes. The ferry can carry a maximum of three cars on each trip.

(b) One day all the cars waiting at $13:00$ get on the ferry. Find the probability that all the cars that arrive in the next $20$ minutes will get on either the $13:10$ or the $13:20$ ferry. [4 marks]

Answer 1

Some hints:

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Note that the probability mass function for: $$X\sim \operatorname*{Po}(\lambda)$$ Is given by: $$P(X=x)=\frac{e^{-\lambda}\cdot \lambda^x}{x!}$$

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For (a), note that there is a mean of $1$ car every $4$ minutes. Therefore, the distribution to consider is: $$X\sim \operatorname*{Po}\left(\frac{1}{4}T\right)$$ Can you now find the value of $T$ from $P(X\leq 3)=0.6$?

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For (b), let $X_1$ and $X_2$ denote the number of cars which arrive in the first $10$ minutes and second period respectively. They must each follow a distribution of: $$X_1\sim \operatorname*{Po}(2.5)$$ $$X_2\sim \operatorname*{Po}(2.5)$$ Now, try finding all the possibilities and add them up to obtain the desired probability.