JEE Advanced 2022 Paper 1 Q16 Mathematics P&C and Probability Probability Hard

JEE Advanced 2022 Paper 1 · Q16 · Probability

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: (I) Probability of $(X_{2}\ge Y_{2})$ is (II) Probability of $(X_{2}>Y_{2})$ is (III) Probability of $(X_{3}=Y_{3})$ is (IV) Probability of $(X_{3}>Y_{3})$ is

List-II: (P) $3/8$ (Q) $11/16$ (R) $5/16$ (S) $355/864$ (T) $77/432$

The correct option is:

  1. A. (I)$\to$(Q); (II)$\to$(R); (III)$\to$(T); (IV)$\to$(S)
  2. B. (I)$\to$(Q); (II)$\to$(R); (III)$\to$(T); (IV)$\to$(T)
  3. C. (I)$\to$(P); (II)$\to$(R); (III)$\to$(Q); (IV)$\to$(S)
  4. D. (I)$\to$(P); (II)$\to$(R); (III)$\to$(Q); (IV)$\to$(T)
Reveal answer + step-by-step solution

Correct answer:A

Solution

Per round: $P(x>y)=P(xY_{2})=(5/12)^{2}+2(5/12)(1/6)=25/144+10/72=25/144+20/144+ tie cases$ — refined: $=5/16$ (R). For round 3: $P(X_{3}=Y_{3})$: enumerate over (W,L,T) outcomes giving equal totals: scoring 5 vs 0 or 2 vs 2. $X_{3}=Y_{3}$ requires all 3 ties OR one tie and one win one loss (in 3! permutations). $P=(1/6)^{3}+3!/(1!1!1!)\cdot(5/12)(5/12)(1/6)=1/216+6\cdot 25/(144\cdot 6)=1/216+25/144=$ common denom 432: $=2/432+75/432=77/432$ (T). $P(X_{3}>Y_{3})$: by symmetry $= (1-77/432)/2=(355/432)/2=355/864$ (S). Match: (I)Q,(II)R,(III)T,(IV)S = option (A).

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