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 t

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. List- I List- II I Probability of X 2 ≥ Y 2 is P 3 8 II Probability of X 2 > Y 2 is Q 11 16 III Probability of X 3 = Y 3 is R 5 16 IV Probability of X 3 > Y 3 is S 355 864 T 77 432 The correct option is:

TestHub · Accepted Answer

Given, P (draw in 1 round) = 6 36 = 1 6 P (win in 1 round) = 1 2 1 - 1 6 = 5 12 P (loss in 1 round) = 5 12 Now finding the probability of all we get, P X 2 > Y 2 = P 10 , 0 + P 7 , 2 = 5 12 × 5 12 + 5 12 × 1 6 × 2 = 45 144 = 5 16 → R P X 2 = Y 2 = P 5 , 5 + P 4 , 4 = 5 12 × 5 12 × 2 + 1 6 × 1 6 = 25 + 2 72 = 3 8 P X 2 ≥ Y 2 = 5 16 + 3 8 = 11 16 → Q P X 3 = Y 3 = P 6 , 6 + P 7 , 7 = 1 6 × 6 × 6 + 5 12 × 1 6 × 5 12 × 6 = 2 432 + 75 432 = 77 432 → T P X 3 > Y 3 = 1 2 1 - 77 432 = 355 864 → S

	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	$(S)$	$\frac{355}{864}$
		$(T)$	$\frac{77}{432}$

Mathematics - Probability Question with Solution | TestHub

Options:

Doubts & Discussion