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]
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projectilemotion
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Mr T
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2Hi, and welcome to Math.SE! What have you tried to solve the problem? It helps us know exactly where you are stuck, so we can help in the best way possible (We then don't repeat information you already know). – projectilemotion Mar 28 '17 at 09:55
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Some hints:
Note that the probability mass function for: $$X\sim \operatorname*{Po}(\lambda)$$ Is given by: $$P(X=x)=\frac{e^{-\lambda}\cdot \lambda^x}{x!}$$
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 $P(X\leq 3)=0.6$?
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.
projectilemotion
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