JEE Advanced 2022 · Paper 1 · Mathematics Q16

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MathematicsList match+3 / -1Question 16 of 54

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 $i^{th}$ round.

The correct option is:

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
$\frac38$
Q
$\frac{11}{16}$
R
$\frac5{16}$
S
$\frac{355}{864}$
T
$\frac{77}{432}$

(I) → (Q); (II) → (R); (III) → (T); (IV) → (S)

(I) → (Q); (II) → (R); (III) → (T); (IV) → (T)

(I) → (P); (II) → (R); (III) → (Q); (IV) → (S)

(I) → (P); (II) → (R); (III) → (Q); (IV) → (T)

Answer and worked solution

Official answer: A

One round. Let $D=X-Y$ change by $+5$ , $0$ or $-5$ , with probabilities
$P(x>y)=\tfrac{15}{36}=\tfrac5{12},\quad P(x=y)=\tfrac16,\quad P(x<y)=\tfrac5{12}.$

Two rounds.

$P(X_2>Y_2)=\left(\frac5{12}\right)^2+2\cdot\frac5{12}\cdot\frac16=\frac{45}{144}=\frac5{16}$ → (II) R.
$P(X_2=Y_2)=\frac1{36}+2\left(\frac5{12}\right)^2=\frac{54}{144}$ , so $P(X_2\ge Y_2)=\frac{99}{144}=\frac{11}{16}$ → (I) Q.

Three rounds.

$P(X_3=Y_3)=\left(\frac16\right)^3+3!\cdot\frac5{12}\cdot\frac5{12}\cdot\frac16=\frac{4+150}{864}=\frac{77}{432}$ → (III) T.
By symmetry, $P(X_3>Y_3)=\frac12\left(1-\frac{77}{432}\right)=\frac{355}{864}$ → (IV) S.

Topics:Independent events Random variables and distributions

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JEE Advanced 2022 · Paper 1, Mathematics question 16