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 face of the die have six distinct numbers. Let x and y denote the readings on the die rolled by $P_{1}$ and $P_{2}$ respectively If $x > y$ , then $P_{1}$ scores 5 points and $P_{2}$ scores 0 point . If $x = y,$ then each player scores 2 points If $x < y,$ then $P_{1}$ scores 0 point and $P_{2}$ scroes 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 $i^{t h}$ round.

List – I

List – II
I
Probability of $(X_{2} \underline{>} Y_{2})$ is
P
$\frac{3}{8}$
II
Probability of $(X_{2} > Y_{2})$
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
$\frac{77}{432}$

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

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

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

$(I) \to (Q); (I I) \to (R); (I I I) \to (T); (I V) \to (T)$

answer is A.

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Detailed Solution

$P (draw in any round) = \frac{6}{36} = \frac{1}{6}$

$P (win in any round) = P (loss in any round) = \frac{1}{2} (1 - \frac{1}{6}) = \frac{5}{12}$

$P (X_{2} = Y_{2}) = P (5, 5) + P (4, 4) = \frac{5}{12} \times \frac{5}{12} \times 2 + \frac{1}{6} \times \frac{1}{6} = \frac{27}{72} = \frac{3}{8} P (X_{2} > Y_{2}) = \frac{1}{2} (1 - \frac{3}{8}) = \frac{5}{16} P (X_{3} = Y_{3}) = P (6, 6) + P (7, 7) = \frac{1}{6 \times 6 \times 6} + \frac{5}{12} \times \frac{1}{6} \times \frac{5}{12} \times 3! = \frac{2}{432} + \frac{75}{432} = \frac{77}{432}$

$P (X_{3} > Y_{3}) = \frac{1}{2} (1 - \frac{77}{432}) = \frac{355}{864}$

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	List – I		List – II
I	Probability of $(X_{2} \underline{>} Y_{2})$ is	P	$\frac{3}{8}$
II	Probability of $(X_{2} > Y_{2})$	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	$\frac{77}{432}$