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Question
A car hiring firm has two cars. The demand for cars on each day is distributed as a Poison variate, with mean 1.5. Calculate the proportion of days on which neither car is used
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Solution
In a posion distribution n=2
Mean λ = 1.5
x follows poison distribution
With in P(x) = `("e"^(-lambda) lambda^x)/(x!)`
P(neither car is used) = P(X = 0)
`("e"^(-1.5) (1.5)^0)/(0!) = "e"^(-1.5)`
= 0.2231
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