Two players, P₁ and P₂, 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 P₁ and P₂,respectively. If xy, then P₁ scores 5 points and P₂ scores 0 point. If x=y, then each player scores 2 points. If x List-I List-II (I) Probability of (X₂ ≥ Y₂) is (P) 3/8 (II) Probability of (X₂ Y₂) is (Q) 11/16 (III) Probability of (X₃=Y₃) is (R) 5/16 (IV) Probability of (X₃ Y₃) is (S) 355/864 (T) 77/432 [JEE Adv 2022 P1]

Question

Two players, P₁ and P₂, 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 P₁ and P₂,respectively. If x>y, then P₁ scores 5 points and P₂ scores 0 point. If x=y, then each player scores 2 points. If x List-I List-II (I) Probability of (X₂ ≥ Y₂) is (P) 3/8 (II) Probability of (X₂ > Y₂) is (Q) 11/16 (III) Probability of (X₃=Y₃) is (R) 5/16 (IV) Probability of (X₃ > Y₃) is (S) 355/864 (T) 77/432 [JEE Adv 2022 P1]

Examsnet · Accepted Answer

Correct Answer is : (I) →arrow→ (Q); (II) →arrow→ (R); (III) →arrow→ (T); (IV) →arrow→ (S)

	List-I		List-II
(I)	Probability of $(X_2 \ge Y_2)$ 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)	$\frac{77}{432}$

JEE Advanced 2022 Paper 1