A random variable X follows a Poisson distribution with mean 0.8. Match List-I with List-II (values correct to three decimal places; given \(e^{-0.8} = 0.4493\))
| List-I | List-II | ||
| (A) | \(P(X = 0)\) | (I) | 0.144 |
| (B) | \(P(X = 1)\) | (II) | 0.551 |
| (C) | \(P(X = 2)\) | (III) | 0.449 |
| (D) | \(P(X \ge 1)\) | (IV) | 0.359 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
With \(m = 0.8\), the Poisson probability is \(P(X = r) = \frac{e^{-m}\,m^{r}}{r!}\).
(A) \(P(X = 0)\)
This matches (III) 0.449.
\(P(X = 0) = e^{-0.8} = 0.4493\)
(B) \(P(X = 1)\)
This matches (IV) 0.359.
\(P(X = 1) = 0.4493 \times 0.8 = 0.3594\)
(C) \(P(X = 2)\)
This matches (I) 0.144.
\( \displaystyle P(X = 2) = \frac{0.4493 \times 0.64}{2} = 0.1438 \)
(D) \(P(X \ge 1)\)
This matches (II) 0.551. It is the complement of X = 0:
\(P(X \ge 1) = 1 - 0.4493 = 0.5507\)