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 < y, then P₁ scores 0 point and P₂ scores 5 points. Let X_i and Y_i be the total scores of P₁ and P₂, respectively, after playing the i^th round.
 

List-I		List-II
I	Probability of (X2 ≥ Y2) is	P	\(\frac{3}{8}\)
II	Probability of (X2 > Y2) is	Q	\(\frac{11}{16}\)
III	Probability of (X3 = Y3) is	R	\(\frac{5}{16}\)
IV	Probability of (X3 > Y3) is	S	\(\frac{355}{864}\)
		T	\(\frac{77}{432}\)

The correct option is:

• A. (I) → (Q); (II) → (R); (III) → (T); (IV) → (S)
• B. (I) → (Q); (II) → (R); (III) → (T); (IV) → (T)
• C. (I) → (P); (II) → (R); (III) → (Q); (IV) → (S)
• D. (I) → (P); (II) → (R); (III) → (Q); (IV) → (T)

Answer 1

The Correct Option is A

Let A1; P1 won the round → \( P(A1) = \frac{6C2}{6^2} = \frac{6 \times 5}{6 \times 6 \times 6} = \frac{5}{12} \)

Let A2; P2 won the round → \( P(A2) = \frac{5}{12} \)

Let D; round ends in draw → \( P(D) = \frac{6}{6^2} = \frac{1}{6} \)

(i) \( P(X_2 \geq Y_2) = P(A1 \cap A1) + 2P(A1 \cap A1) + P(D \cap D) + 2P(A1 \cap D) \)

\[ P(X_2 \geq Y_2) = \frac{5}{12} \times \frac{5}{12} + \frac{5}{12} \times \frac{1}{6} + \frac{5}{12} \times \frac{1}{6} + \frac{1}{6} = \frac{11}{16} \]

⇒ (1) → (Q)

(ii) \( P(X_2 \geq Y_2) = P(A1 \cap A1) + 2P(A1 \cap A1) \)

\[ P(X_2 \geq Y_2) = \frac{5 \times 5}{12 \times 12} + 2 \times \frac{25}{144} + \frac{45}{144} + \frac{36}{144} \]

⇒ (ii) → (R)

(iii) \( P(X_3 \geq Y_3) = 3P(A1 \cap A1) + P(D \cap D) \)

\[ P(X_3 \geq Y_3) = 6 \times \frac{5}{12} \times \frac{5}{12} \times \frac{1}{6} \times 6 = \frac{25}{144} + \frac{75}{72} + \frac{432}{432} \]

⇒ (iii) → (T)

(iv) \( P(X_3 \geq Y_3) = P(A1 \cap A1) + P(A2 \cap A2) \)

\[ P(X_3 \geq Y_3) = \frac{5}{12} \times \frac{5}{12} + 3 \times P(A1 \cap D) + P(D \cap D) \]

⇒ IV → (S)

Answer 2

Let \( A_1 \); \( P_1 \) won the round

\(\Rightarrow P(A_1) = \frac{6C2}{6^2} = \frac{6 \times 5}{6 \times 6 \times 6} = \frac{5}{12} \)

Let \( A_2 \); \( P_2 \) won the round \(\Rightarrow P(A_2) = \frac{5}{12} \)

Let \( D \); round ends in draw \(\Rightarrow P(D) = \frac{6}{6^2} = \frac{1}{6} \)

(i) \( P(X_2 \geq Y_2) = P(A_1 \cap A_1) + 2P(A_1 \cap A_1) + P(D \cap D) + 2P(A_1 \cap D) \) \[ P(X_2 \geq Y_2) = \frac{5}{12} \times \frac{5}{12} + \frac{5}{12} \times \frac{1}{6} + \frac{5}{12} \times \frac{1}{6} + \frac{1}{6} = \frac{11}{16} \] \(\Rightarrow (1) \rightarrow (Q)\) \textbf{(ii)} \( P(X_2 \geq Y_2) = P(A_1 \cap A_1) + 2P(A_1 \cap A_1) \) \[ P(X_2 \geq Y_2) = \frac{5 \times 5}{12 \times 12} + 2 \times \frac{25}{144} + \frac{45}{144} + \frac{36}{144} \] \(\Rightarrow (ii) \rightarrow (R)\) \textbf{(iii)} \( P(X_3 \geq Y_3) = 3P(A_1 \cap A_1) + P(D \cap D) \) \[ P(X_3 \geq Y_3) = 6 \times \frac{5}{12} \times \frac{5}{12} \times \frac{1}{6} \times 6 = \frac{25}{144} + \frac{75}{72} + \frac{432}{432} \] \(\Rightarrow (iii) \rightarrow (T)\) \textbf{(iv)} \( P(X_3 \geq Y_3) = P(A_1 \cap A_1) + P(A_1 \cap A_1) + 3P(A_1 \cap A_1) + 3P(D \cap D) \) \[ P(X_3 \geq Y_3) = \frac{5}{12} \times \frac{5}{12} + 3 \times P(A_1 \cap D) + 3 \times P(A_1 \cap D) + P(D \cap D) \] \(\Rightarrow IV \rightarrow (S)\)