Two players, and , 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 and denote the readings on the die rolled by and , respectively. If , then

Question

Two players, $P_{1}$ and $P_{2}$ , 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 $P_{1}$ 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 $P_{2}$ scores 5 points. Let $X_{i}$ and $Y_{i}$ be the total scores of $P_{1}$ and $P_{2}$ , respectively, after playing the $i^{th}$ round. [2022]

	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 $(X_{3} > Y_{3})$	(S)	$\frac{355}{864}$
		(T)	$\frac{77}{432}$

The correct option is:

	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 $(X_{3} > Y_{3})$	(S)	$\frac{355}{864}$
		(T)	$\frac{77}{432}$

The correct option is:

Smart Achievers · Accepted Answer

(1)

$P (X_{i} > Y_{i}) + P (X_{i} < Y_{i}) + P (X_{i} = Y_{i}) = 1$

$and P (X_{i} > Y_{i}) = P (X_{i} < Y_{i}) = p$

$For i = 2$

$P (X_{2} = Y_{2}) = P (5, 5) + P (4, 4)$

$= \frac{5}{12} \times \frac{5}{12} \times 2 + \frac{1}{6} \times \frac{1}{6}$ $= \frac{25}{72} + \frac{1}{36} = \frac{27}{72} = \frac{3}{8}$

$P (X_{2} > Y_{2}) = P (10, 0) = \frac{5}{12} \times \frac{5}{12} + \frac{5}{12} \times \frac{1}{6} \times 2 = \frac{5}{16}$

$P (X_{2} \geq Y_{2}) = \frac{5}{16} + \frac{3}{8} = \frac{11}{16}$

$I \to Q, I I \to R$

$For i = 3$

$P (X_{3} = Y_{3}) = P (6, 6) + P (7, 7)$

$= \frac{1}{6 \times 6} + \frac{5}{12} \times \frac{1}{6} \times \frac{5}{12} \times 6 = \frac{77}{432}$

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

$I I I \to T, I V \to S$

Solution

1 (I) → (Q); (II) → (R); (III) → (T); (IV) → (S)

2 (I) → (Q); (II) → (R); (III) → (T); (IV) → (T)

3 (I) → (P); (II) → (R); (III) → (Q); (IV) → (S)

4 (I) → (P); (II) → (R); (III) → (Q); (IV) → (T)

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