Probability: P(X>Y) is | JEE Advanced 2016 Paper 2

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

Football teams $T_1$ and $T_2$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T_1$ winning, drawing and losing a game against $T_2$ are $\frac{1}{2}, \frac{1}{6}$ and $\frac{1}{3}$, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let $X$ and $Y$ denote the total points scored by teams $T_1$ and $T_2$, respectively, after two games.

JEE Advanced 2016 Paper 2Question 1

$P(X>Y)$ is

Probability

$\frac{1}{4}$$\frac{5}{12}$✓ Correct answer$\frac{1}{2}$$\frac{7}{12}$

Solution

$X>Y$ for: WW $\frac14$, WD $2\cdot\frac1{12}=\frac16$, WL $0$ (tie 3-3), DD tie, DL $2\cdot\frac1{18}$ gives $Y$ larger. So $P(X>Y)=\frac14+\frac16=\frac5{12}$.

Probability - JEE Mathematics question (JEE Advanced 2016 Paper 2)

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