Verified Solution • 2025–26 Session moderate

QJEEAdv22Ma74b9: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Ma74b9 Solution: Two players, and , play a game against each other. In every…

Two players, P1P_{1} and P2P_{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 xx and yy denote the readings on the die rolled by P1P_{1} and P2P_{2} , respectively. If x>y , then P1P_{1} scores 5 points and P2P_{2} scores 0 point. If x=yx=y , then each player scores 2 points. If x < y , then P1P_{1} scores 0 point and P2P_{2} scores 5 points. Let XiX_{i} and YiY_{i} be the total scores of P1P_{1} and P2P_{2} , respectively, after playing the ith i^{\text {th }} round.

List-I List-II
(I) Probability of (X2≥Y2)\left(X_{2} \geq Y_{2}\right) is (P) 38\frac{3}{8}
(II) Probability of \left(X_{2}>Y_{2}\right) is (Q) 1116\frac{11}{16}
(III) Probability of (X3=Y3)\left(X_{3}=Y_{3}\right) is (R) 516\frac{5}{16}
(IV) Probability of \left(X_{3}>Y_{3}\right) is (S) 355864\frac{355}{864}
(T) 77432\frac{77}{432}

The correct option is:

Multiple Choice Options

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

Step-by-Step Verified Solution

Given,

amp;P( draw in 1 round )=636=16amp;P( win in 1 round )=12(1−16)=512amp;P( loss in 1 round )=512\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}

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