Let X denotes the number of doublets obtained in 3 throws of a pair of dice.
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(P(X = 0)\) | (I) | \(\displaystyle\frac{1}{216}\) |
| (B) | \(P(X=1)\) | (II) | \(\displaystyle\frac{15}{216}\) |
| (C) | \(P(X=2)\) | (III) | \(\displaystyle\frac{75}{216}\) |
| (D) | \(P(X = 3)\) | (IV) | \(\displaystyle\frac{125}{216}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
A throw of a pair of dice gives a doublet (both dice show the same number) in 6 of the 36 outcomes:
\( \displaystyle p = \frac{6}{36} = \frac{1}{6},\quad q = 1 - p = \frac{5}{6} \)
The 3 throws are independent, so X follows the binomial distribution with n = 3:
\( P(X = r) = {}^{3}C_{r}\, p^{r} q^{3-r} \)
(A) \(P(X = 0)\)
This matches (IV):
\( \displaystyle P(X = 0) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \)
(B) \(P(X=1)\)
This matches (III):
\( \displaystyle P(X = 1) = 3 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^2 = \frac{75}{216} \)
(C) \(P(X=2)\)
This matches (II):
\( \displaystyle P(X = 2) = 3 \times \left(\frac{1}{6}\right)^2 \times \frac{5}{6} = \frac{15}{216} \)
(D) \(P(X = 3)\)
This matches (I):
\( \displaystyle P(X = 3) = \left(\frac{1}{6}\right)^3 = \frac{1}{216} \)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).