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JEE Advanced 2016
Paper-2 2016
Question
Football teams T1 and T2 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 drawn and 0 point for a loss in a game. Let X and Y denote the total points secored by teams $\mathrm{T}_1$ and $T_2$, respectively, after two games.

P(X > Y) is
Select the correct option:
A
$\frac{1}{4}$
B
$\frac{5}{12}$
C
$\frac{1}{2}$
D
$\frac{7}{12}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & P\left(T_{1 \mathrm{w}}\right)=\frac{1}{2}, P\left(T_{1 \mathrm{D}}\right)=\frac{1}{6}, P\left(T_{1 \mathrm{~L}}\right)=\frac{1}{3} \\ & 3 \\ & x>y \Rightarrow \frac{1 \mid 2}{T_1} \\ & P(W, W D)=\frac{1}{2} \times\left(\frac{1}{2}+\frac{1}{6}\right)\end{aligned}$
$\begin{aligned} & =\frac{1}{2} \times\left(\frac{6+2}{12}\right) \\ & =\frac{8}{24}=\frac{1}{3} \\ & P(D \text { and } W)=\frac{1}{6} \times \frac{1}{2}=\frac{1}{12} \\ & \Rightarrow P(x>y)=\frac{1}{3}+\frac{1}{12}=\frac{4}{12}+\frac{1}{12}=\frac{5}{12} \Rightarrow \mathbf{B}\end{aligned}$
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