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 $X_l$ and $Y_l$ be the total scores of $P_1$ and $P_2$, respectively, after playing the $l^{\text{th}}$ 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:

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 Xₗ and Yₗ be the total scores of P₁ and P₂, respectively, after playing the lᵗᵉˣ

$x>y \Rightarrow P_1:5,\ P_2:0$

$x=y \Rightarrow 2 \text{ each}$

$x<y \Rightarrow P_1:0,\ P_2:5$

$\text{To find: the correct matching option}$

$P(x>y) = \dfrac{15}{36} = \dfrac{5}{12} \ (=p)$

$P(x=y) = \dfrac{6}{36} = \dfrac{1}{6} \ (=r)$

$P(x<y) = \dfrac{15}{36} = \dfrac{5}{12} \ (=q)$

$D_l = X_l - Y_l = \delta_1 + \delta_2 + \cdots + \delta_l$

$\delta \in \{+5,\ 0,\ -5\}$

$P(D_2=10)=p^2=\dfrac{25}{144}$

$P(D_2=5)=2pr=\dfrac{20}{144},\quad P(D_2=0)=r^2+2pq=\dfrac{54}{144}$

$P(X_2 \ge Y_2) = P(D_2 \ge 0) = \dfrac{99}{144} = \dfrac{11}{16} \ \to (Q)$

$P(X_2 > Y_2) = P(D_2 > 0) = \dfrac{45}{144} = \dfrac{5}{16} \ \to (R)$

P(D₃=0) = r³ + 6pqr = 8/1728+300/1728

$= \dfrac{308}{1728} = \dfrac{77}{432} \ \to (T)$

$P(D_3>0) = P(D_3<0) = \dfrac{1 - 308/1728}{2}$

$= \dfrac{710}{1728} = \dfrac{355}{864} \ \to (S)$

I → Q, II → R, III → T, IV → S

$\boxed{\text{Answer: option (1)}}$