Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice.
Match List-I with List-II
|
List-I X |
List-II Probability, P(X) |
|
(A) 4 |
(I) $\frac{1}{6}$ |
|
(B) 5 |
(II) $\frac{5}{36}$ |
|
(C) 6 |
(III) $\frac{1}{12}$ |
|
(D) 7 |
(IV) $\frac{1}{9}$ |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
|
List-I X |
List-II Probability, P(X) |
|
(A) 4 |
(III) $\frac{1}{12}$ |
|
(B) 5 |
(IV) $\frac{1}{9}$ |
|
(C) 6 |
(II) $\frac{5}{36}$ |
|
(D) 7 |
(I) $\frac{1}{6}$ |
For two fair dice, total outcomes = $36$.
$X=4$: outcomes $(1,3),(2,2),(3,1)$ → $3$ outcomes → $P=\frac{3}{36}=\frac{1}{12}$ → (III).
$X=5$: outcomes $(1,4),(2,3),(3,2),(4,1)$ → $4$ outcomes → $P=\frac{4}{36}=\frac{1}{9}$ → (IV).
$X=6$: outcomes $(1,5),(2,4),(3,3),(4,2),(5,1)$ → $5$ outcomes → $P=\frac{5}{36}$ → (II).
$X=7$: outcomes $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ → $6$ outcomes → $P=\frac{6}{36}=\frac{1}{6}$ → (I).
Final answer: (A)–(III), (B)–(IV), (C)–(II), (D)–(I)