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 $i^{\text{th }}$ round.

List I		List II
I	Probability of $\left(X_2\ge Y_2\right)$ is	P	$\frac{3}{8}$
II	Probability of $\left(X_2>Y_2\right)$ is	Q	$\frac{11}{16}$
III	Probability of $\left(X_3=Y_3\right)$ is	R	$\frac{5}{16}$
IV	Probability of $\left(X_3>Y_3\right)$ 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)

Answer 1

Answer: (A)

$P(\text{ draw in }1\text{ round })=\frac{6}{36}=\frac{1}{6}$
$P(\text{ win in }1\text{ round })=\frac{1}{2}\left(1-\frac{1}{6}\right)=\frac{5}{12}$
$P(\text{ loss in }1\text{ round })=\frac{5}{12}$
$P\left(X_2>Y_2\right)=P(10,0)+P(7,2)=\frac{5}{12}\times \frac{5}{12}+\frac{5}{12}\times \frac{1}{6}\times 2=\frac{45}{144}=\frac{5}{16}$
$P\left(X_2=Y_2\right)=P(5,5)+P(4,4)=\frac{5}{12}\times \frac{5}{12}\times 2+\frac{1}{6}\times \frac{1}{6}=\frac{25+2}{72}=\frac{3}{8}$
$P\left(X_3=Y_3\right)=P(6,6)+P(7,7)=\frac{1}{6\times 6\times 6}+\frac{5}{12}\times \frac{1}{6}\times \frac{5}{12}\times 6=\frac{2}{432}+\frac{75}{432}=\frac{77}{432}$
$P\left(X_3>Y_3\right)=\frac{1}{2}\left(1-\frac{77}{432}\right)=\frac{355}{864}$