JEE Advanced 2022 — Paper 1 — MATH-III solution

JEE Advanced2022Paper 1MATH-III

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.

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:

A
(I) $\to$ (Q); (II) $\to$ (R); (III) $\to$ (T); (IV) $\to$ (S)
B
(I) $\to$ (Q); (II) $\to$ (R); (III) $\to$ (T); (IV) $\to$ (T)
C
(I) $\to$ (P); (II) $\to$ (R); (III) $\to$ (Q); (IV) $\to$ (S)
D
(I) $\to$ (P); (II) $\to$ (R); (III) $\to$ (Q); (IV) $\to$ (T)

Solution

In a single round, $P (x > y) = P (x < y) = \frac{15}{36} = \frac{5}{12}$ and $P (x = y) = \frac{6}{36} = \frac{1}{6}$ . Denote $p_{W} = \frac{5}{12}$ (win), $p_{D} = \frac{1}{6}$ (draw). By symmetry $P (X_{i} > Y_{i}) = P (X_{i} < Y_{i})$ for every $i$ .

After 2 rounds. Each round contributes (+5,0), (+2,+2) or (0,+5) to scores. $X_{2} = Y_{2}$ happens iff both rounds are draws OR one round is a $P_{1}$ -win and the other a $P_{2}$ -win.

$P (X_{2} = Y_{2}) = p_{D}^{2} + 2 p_{W}^{2} = \frac{1}{36} + 2 \cdot \frac{25}{144} = \frac{4}{144} + \frac{50}{144} = \frac{54}{144} = \frac{3}{8} .$

$P (X_{2} > Y_{2}) = \frac{1 - P ( X _{2} = Y _{2} )}{2} = \frac{1}{2} (1 - \frac{3}{8}) = \frac{5}{16} .$

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

$\Rightarrow$ (I) $\to$ (Q), (II) $\to$ (R).

After 3 rounds. $X_{3} = Y_{3}$ requires either all three draws, or one of each (one $P_{1}$ -win, one $P_{2}$ -win, one draw):

$P (X_{3} = Y_{3}) = p_{D}^{3} + 3! p_{W}^{2} p_{D} = \frac{1}{216} + 6 \cdot \frac{25}{144} \cdot \frac{1}{6} = \frac{1}{216} + \frac{25}{144} .$

Common denominator $432$ : $\frac{1}{216} = \frac{2}{432}$ , $\frac{25}{144} = \frac{75}{432}$ . So $P (X_{3} = Y_{3}) = \frac{77}{432}$ .

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

$\Rightarrow$ (III) $\to$ (T), (IV) $\to$ (S).

Matching: (I) $\to$ (Q); (II) $\to$ (R); (III) $\to$ (T); (IV) $\to$ (S), which is option (A).

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