A cab hire firm has two cabs, which it hires out day by day. The number of demands for cabs on each day is distributed as a Poisson distribution with mean of 1.5, then the probability of days on which neither cab is used is (Use \( e^{-1.5} \) = 0.2231)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.2231
Neither cab is used on a day when there is no demand, that is, X = 0, where X is the Poisson number of demands with mean 1.5:
\( \displaystyle P(X=0)=\frac{e^{-1.5}(1.5)^0}{0!} \)
\( \displaystyle =e^{-1.5}=0.2231 \)