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

List I		List II
I	Probability of $\left(X_2 \geq Y_2\right)$ 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	$\frac{77}{432}$

• A. (I) $\rightarrow$ (Q); (II) $\rightarrow$ (R); (III) $\rightarrow$ (T); (IV) $\rightarrow$ (S)
• B. (I) $\rightarrow$ (Q); (II) $\rightarrow$ (R); (III) $\rightarrow$ (T); (IV) $\rightarrow$ (T)
• C. (I) $\rightarrow$ (P); (II) $\rightarrow$ (R); (III) $\rightarrow$ (Q); (IV) $\rightarrow$ (S)
• D. (I) $\rightarrow$ (P); (II) $\rightarrow$ (R); (III) $\rightarrow$ (Q); (IV) $\rightarrow$ (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}$