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:

1. Find the Total Number of Outcomes:

Two numbers are chosen without replacement from the first six positive integers $\{1, 2, 3, 4, 5, 6\}$.

The total number of ways to pick a pair of distinct numbers is:

$\text{Total Outcomes} = \binom{6}{2} = \frac{6 \times 5}{2} = 15$

2. Understand the Random Variable $x$:

Since $x$ denotes the larger of the two selected numbers, if $x = k$, then the other selected number must be strictly less than $k$ (chosen from the $k-1$ available smaller numbers).

$\text{Number of favorable pairs for } x = k \text{ is } (k - 1)$

3. Calculate the Probabilities for List-I:

• For (A) $P(x = 2)$:

Favorable pair: $(1, 2)$ $\implies 1$ way.

$P(x = 2) = \frac{1}{15} \quad \rightarrow \mathbf{(II)}$

• For (B) $P(x = 3)$:

Favorable pairs: $(1, 3), (2, 3)$ $\implies 2$ ways.

$P(x = 3) = \frac{2}{15} \quad \rightarrow \mathbf{(III)}$

• For (C) $P(x = 4)$:

Favorable pairs: $(1, 4), (2, 4), (3, 4)$ $\implies 3$ ways.

$P(x = 4) = \frac{3}{15} = \frac{1}{5} \quad \rightarrow \mathbf{(IV)}$

• For (D) $P(x = 5)$:

Favorable pairs: $(1, 5), (2, 5), (3, 5), (4, 5)$ $\implies 4$ ways.

$P(x = 5) = \frac{4}{15} \quad \rightarrow \mathbf{(I)}$