A car hire firm has two cars, which is hire out day by day. The number of demands for cars on each day is distributed as a Poisson distribution with mean 1.5. Then which of the following is/are TRUE? (use \( e^{-1.5}=0.2231 \))
A. The probability of days when no car is required is 0.2231
B. The probability of days when some demand is refused is 0.1913
C. The probability of days when no car is required is 0.1913
D. The probability of days when some demand is refused is 0.2231
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A and B only
Let X be the number of cars demanded on a day; X follows a Poisson distribution with mean 1.5. No car is required when X = 0. The firm has only two cars, so some demand is refused when X > 2.
\( P(X=0)=e^{-1.5}=0.2231 \)
\( \displaystyle P(X\le 2)=e^{-1.5}\left(1+1.5+\frac{1.5^2}{2}\right) \)
\( =0.2231\times 3.625 \)
\( =0.8087 \)
\( P(X\gt 2)=1-0.8087=0.1913 \)
So the probability that no car is required is 0.2231, and the probability that some demand is refused is 0.1913.
Checking the statements
(A) The probability of days when no car is required is 0.2231 — Correct. \( P(X=0)=e^{-1.5}=0.2231 \).
(B) The probability of days when some demand is refused is 0.1913 — Correct. \( P(X\gt 2)=1-0.8087=0.1913 \).
(C) The probability of days when no car is required is 0.1913 — Incorrect. That probability is 0.2231, not 0.1913.
(D) The probability of days when some demand is refused is 0.2231 — Incorrect. That probability is 0.1913, not 0.2231.