The number of customers who enter a small shop in a 10-minute interval follows a Poisson distribution with mean 4. The probability that at most 2 customers enter the shop in a given 10-minute interval is (Given \(e^{-4} = 0.0183\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.2379
Let X be the number of customers in the interval. X follows a Poisson distribution with \(m = 4\):
\( \displaystyle P(X = r) = \frac{e^{-m}\,m^{r}}{r!} \)
At most 2 means X = 0, 1 or 2:
\( \displaystyle P(X \le 2) = e^{-4}\left(1 + 4 + \frac{4^{2}}{2!}\right) \)
\(= 0.0183 \times (1 + 4 + 8)\)
\(= 0.0183 \times 13 = 0.2379\)
Why the other options are incorrect
Option (1) 0.0915 — Incorrect. This is \(0.0183 \times 5\), the probability of at most 1 customer.
Option (2) 0.7621 — Incorrect. This is 1 − 0.2379, the probability of more than 2 customers.
Option (3) 0.1464 — Incorrect. This is \(0.0183 \times 8\), the probability of exactly 2 customers.