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Two players, $P_{1} a n d P_{2},$ play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let x and y denote the readings on the die rolled by $P_{1} a n d P_{2},$ respectively. If x > y, then P₁ scores 5 points and $P_{2}$ scores 0 point. If x = y, then each player score 2 points. If x < y, then P₁ scores 0 point and $P_{2}$ scores 5 points. Let $X_{i} a n d Y_{i}$ be the total scores of $P_{1} a n d P_{2},$ respectively, after playing the i^th round.
List – I List – II
I) Probability of $(X_{2} \geq Y_{2})$ is P) $\frac{3}{8}$
II) Probability of $(X_{2} > Y_{2})$ is Q) $\frac{11}{16}$
III) Probability of $(X_{3} = Y_{3})$ is R) $\frac{5}{16}$
IV) Probability of $(X_{3} > Y_{3})$ is S) $\frac{355}{864}$
T) $\frac{77}{432}$
The correct option is

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$I \to Q; I I \to R; I I I \to T; I V \to P$

$I \to P; I I \to R; I I I \to Q; I V \to S$

$I \to Q; I I \to R; I I I \to T; I V \to S$

$I \to P; I I \to R; I I I \to Q; I V \to T$

answer is A.

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Detailed Solution

(1)
$\begin{array}{l} S = {(1, 1), (1, 2), (1, 3) ..., (1, 6) \\ \underset{\begin{array}{l} . \\ . \\ . \end{array}}{(2, 1)}, (2, 2), (2, 3) ..., (2, 6) \\ (6, 1), (6, 2), (6, 3) ..., (6, 6)} \end{array}$
$P (P_{1}) = \frac{5}{12} = P (P_{2}), P (D) = \frac{1}{6} i) P (X_{2} \geq Y_{2}) = P (P_{1}) . P (P_{1}) + 2 P (P_{1}) . P (P_{1}) + P (D) . P (D) + 2 P (P_{1}) . P (D) = \frac{5}{12} \times \frac{5}{12} + 2 (\frac{5}{12}) (\frac{5}{12}) + \frac{1}{6} \times \frac{1}{6} + 2. \frac{5}{12} . \frac{1}{6} = \frac{11}{16} i i) P (X_{2} > Y_{2}) = P (P_{1}) . P (P_{1}) + 2 P (P_{1}) . P (D) = \frac{5}{16} i i i) P (X_{3} = Y_{3}) = 3 P (P_{1}) . P (D) . P (P_{2}) + P (D) . P (D) . P (D) = \frac{77}{432} i v) P (X_{3} > Y_{3}) = P (P_{1}) . P (P_{1}) . P (P_{1}) + 3 P (P_{1}) . P (P_{1}) . P (P_{2}) + 3 P (P_{1}) . P (P_{1}) . P (D) + 3 P (P_{1}) . P (D) . P (D) = \frac{355}{864}$

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