Weekly supply to car dealer B follows a Poisson distribution with mean $\frac{2}{3}$ cars. What is the probability that, in a given week, fewer than three cars are supplied to B?

0.856

0.920

0.970

0.995

Explanation: For

X\sim\text{Po}(2/3)

P(X<3) = P(0)+P(1)+P(2) = e^{-2/3}\left[1+\frac23+\frac{(2/3)^2}{2}\right] = e^{-2/3}(1.8889) \approx 0.970

. Omitting the

P(2)

term gives

e^{-2/3}(1.6667)\approx0.856

. Rounding the mean up to 1 before calculating gives

e^{-1}(2.5)\approx0.920

. Including

P(3)

as well (an off-by-one error using

X\leq3

) gives

\approx0.995

Derived from ZIMSEC Maths Paper 1, June 2007, Q6 (Statistics)

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