Two numbers are selected without replacement at random, one at a time from the first six positive integers. Let x denotes the larger of the two numbers.
Match List-I with List-II
|
List-I |
List-II |
|
(A) $P(x = 2)$ |
(I) $\frac{4}{15}$ |
|
(B) $P(x = 3)$ |
(II) $\frac{1}{15}$ |
|
(C) $P(x = 4)$ |
(III) $\frac{2}{15}$ |
|
(D) $P(x = 5)$ |
(IV) $\frac{1}{5}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
|
List-I |
List-II |
|
(A) $P(x = 2)$ |
(II) $\frac{1}{15}$ |
|
(B) $P(x = 3)$ |
(III) $\frac{2}{15}$ |
|
(C) $P(x = 4)$ |
(IV) $\frac{1}{5}$ |
|
(D) $P(x = 5)$ |
(I) $\frac{4}{15}$ |
The first six positive integers are: 1, 2, 3, 4, 5, 6.
Total number of ways to choose 2 numbers (without replacement):
6C2 = 15
Let $x$ denote the larger of the two selected numbers.
(A) $P(x = 2)$:
Only one pair possible: (1,2) → larger = 2
So, $P(x=2) = \frac{1}{15}$
(B) $P(x = 3)$:
Valid pairs: (1,3), (2,3) → 2 pairs
$P(x=3) = \frac{2}{15}$
(C) $P(x = 4)$:
Valid pairs: (1,4), (2,4), (3,4) → 3 pairs
$P(x=4) = \frac{3}{15} = \frac{1}{5}$
(D) $P(x = 5)$:
Valid pairs: (1,5), (2,5), (3,5), (4,5) → 4 pairs
$P(x=5) = \frac{4}{15}$