JEE Advance - Mathematics (2022 - Paper 1 Online - No. 16)
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 $$\left(X_{2} \geq Y_{2}\right)$$ 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) $$\frac{77}{432}$$ |
The correct option is:
(I) $$\rightarrow$$ (Q); (II) $$\rightarrow$$ (R); (III) $$\rightarrow$$ (T); (IV) $$\rightarrow(S)$$
(I) $$\rightarrow$$ (Q); (II) $$\rightarrow$$ (R); (III) $$\rightarrow$$ (T); (IV) $$\rightarrow$$ (T)
(I) $$\rightarrow$$ (P); (II) $$\rightarrow$$ (R); (III) $$\rightarrow(\mathrm{Q}) ;(\mathrm{IV}) \rightarrow(\mathrm{S})$$
(I) $$\rightarrow$$ (P); (II) $$\rightarrow$$ (R); (III) $$\rightarrow$$ (Q); (IV) $$\rightarrow$$ (T)
Explanation
Given,
$$ \begin{aligned} & P(\text { draw in } 1 \text { round })=\frac{6}{36}=\frac{1}{6} \\\\ & P(\text { win in } 1 \text { round })=\frac{1}{2}\left(1-\frac{1}{6}\right)=\frac{5}{12} \\\\ & P(\text { loss in } 1 \text { round })=\frac{5}{12} \end{aligned} $$
Now finding the probability of all we get.
$$ \begin{aligned} & \mathrm{P}\left(\mathrm{X}_2>\mathrm{Y}_2\right)=\mathrm{P}(10,0)+\mathrm{P}(7,2) \\\\ & =\frac{5}{12} \times \frac{5}{12}+\frac{5}{12} \times \frac{1}{6} \times 2 \\\\ & =\frac{45}{144}=\frac{5}{16} \rightarrow \mathrm{R} \\\\ & \mathrm{P}\left(\mathrm{X}_2=\mathrm{Y}_2\right)=\mathrm{P}(5,5)+\mathrm{P}(4,4) \\\\ & =\frac{5}{12} \times \frac{5}{12} \times 2+\frac{1}{6} \times \frac{1}{6} \\\\ & =\frac{25+2}{72}=\frac{3}{8} \\\\ & \mathrm{P}\left(\mathrm{X}_3=\mathrm{Y}_3\right)=\mathrm{P}(6,6)+\mathrm{P}(7,7) \\\\ & =\frac{1}{6 \times 6 \times 6}+\frac{5}{12} \times \frac{1}{6} \times \frac{5}{12} \times \frac{1}{6} \\\\ & =\frac{77}{432} \rightarrow \mathrm{F} \\\\ & \mathrm{P}\left(\mathrm{X}_3>\mathrm{Y}_3\right)=\frac{1}{2}\left(1-\frac{77}{432}\right)=\frac{355}{864} \rightarrow \mathrm{S} \\\\ & P\left(X_2 \geq Y_2\right)=\frac{5}{16}+\frac{3}{8}=\frac{11}{16} \rightarrow \mathrm{Q} \end{aligned} $$
$$ \begin{aligned} & P(\text { draw in } 1 \text { round })=\frac{6}{36}=\frac{1}{6} \\\\ & P(\text { win in } 1 \text { round })=\frac{1}{2}\left(1-\frac{1}{6}\right)=\frac{5}{12} \\\\ & P(\text { loss in } 1 \text { round })=\frac{5}{12} \end{aligned} $$
Now finding the probability of all we get.
$$ \begin{aligned} & \mathrm{P}\left(\mathrm{X}_2>\mathrm{Y}_2\right)=\mathrm{P}(10,0)+\mathrm{P}(7,2) \\\\ & =\frac{5}{12} \times \frac{5}{12}+\frac{5}{12} \times \frac{1}{6} \times 2 \\\\ & =\frac{45}{144}=\frac{5}{16} \rightarrow \mathrm{R} \\\\ & \mathrm{P}\left(\mathrm{X}_2=\mathrm{Y}_2\right)=\mathrm{P}(5,5)+\mathrm{P}(4,4) \\\\ & =\frac{5}{12} \times \frac{5}{12} \times 2+\frac{1}{6} \times \frac{1}{6} \\\\ & =\frac{25+2}{72}=\frac{3}{8} \\\\ & \mathrm{P}\left(\mathrm{X}_3=\mathrm{Y}_3\right)=\mathrm{P}(6,6)+\mathrm{P}(7,7) \\\\ & =\frac{1}{6 \times 6 \times 6}+\frac{5}{12} \times \frac{1}{6} \times \frac{5}{12} \times \frac{1}{6} \\\\ & =\frac{77}{432} \rightarrow \mathrm{F} \\\\ & \mathrm{P}\left(\mathrm{X}_3>\mathrm{Y}_3\right)=\frac{1}{2}\left(1-\frac{77}{432}\right)=\frac{355}{864} \rightarrow \mathrm{S} \\\\ & P\left(X_2 \geq Y_2\right)=\frac{5}{16}+\frac{3}{8}=\frac{11}{16} \rightarrow \mathrm{Q} \end{aligned} $$
Comments (0)
