The number of goals scored by a football team in a match follows a Poisson distribution with mean 1.5. The probability that the team scores exactly 2 goals in a match is (Given \(e^{-1.5} = 0.2231\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.2510
Let X be the number of goals in a match. X follows a Poisson distribution with \(m = 1.5\):
\( \displaystyle P(X = r) = \frac{e^{-m}\,m^{r}}{r!} \)
For r = 2:
\( \displaystyle P(X = 2) = \frac{e^{-1.5}\,(1.5)^{2}}{2!} \)
\( \displaystyle = \frac{0.2231 \times 2.25}{2} \)
\(= 0.2510\)
Why the other options are incorrect
Option (1) 0.3347 — Incorrect. This is \(P(X = 1) = 0.2231 \times 1.5\), exactly one goal.
Option (3) 0.5020 — Incorrect. This is \(0.2231 \times 2.25\) without dividing by 2! = 2.
Option (4) 0.1255 — Incorrect. This is \(P(X = 3) = \frac{0.2231 \times 3.375}{6}\), exactly three goals.