The number of printing mistakes on a page of a book follows a Poisson distribution with mean 0.5. The probability that a page chosen at random has exactly one printing mistake is (Given \(e^{-0.5} = 0.6065\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.3033
Let X be the number of mistakes on a page. X follows a Poisson distribution with \(m = 0.5\):
\( \displaystyle P(X = r) = \frac{e^{-m}\,m^{r}}{r!} \)
For r = 1:
\( \displaystyle P(X = 1) = \frac{e^{-0.5}\,(0.5)^{1}}{1!} \)
\(= 0.6065 \times 0.5 = 0.30325\)
\(\approx 0.3033\) (to four decimal places)
Why the other options are incorrect
Option (1) 0.6065 — Incorrect. This is \(P(X = 0)\), a page with no mistake.
Option (2) 0.3935 — Incorrect. This is 1 − 0.6065, a page with at least one mistake.
Option (4) 0.0758 — Incorrect. This is \(P(X = 2) = \frac{0.6065 \times 0.25}{2}\), exactly two mistakes.