A telephone switchboard receives, on an average, 2 calls per minute, and the number of calls in a minute follows a Poisson distribution. The probability that the switchboard receives at least 2 calls in a given minute is (Given \(e^{-2} = 0.1353\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.5941
Let X be the number of calls in a minute. X follows a Poisson distribution with m = 2:
\( \displaystyle P(X = r) = \frac{e^{-2}\,2^{r}}{r!} \)
“At least 2” is the complement of X = 0 or X = 1:
\(P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]\)
\(= 1 - (e^{-2} + 2e^{-2})\)
\(= 1 - 3 \times 0.1353\)
\(= 1 - 0.4059 = 0.5941\)
Why the other options are incorrect
Option (1) 0.4059 — Incorrect. This is \(P(X \le 1)\), the probability of fewer than 2 calls.
Option (2) 0.8647 — Incorrect. This is \(1 - P(X = 0)\), the probability of at least 1 call.
Option (4) 0.2706 — Incorrect. This is \(P(X = 2) = \frac{4 \times 0.1353}{2}\), exactly 2 calls only.