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JEE Advanced 2016Paper 2
Exam
JEE Advanced
Year
2016
Paper
Paper 2
Subject
Mathematics
Chapter
Probability

JEE Advanced 2016 · Paper 2 · Mathematics · Probability

Paragraph: Football teams T_ 1 and T_ 2 have to play two games...

Paragraph: Football teams T1T_{1} and T2T_{2} have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1T_{1} winning, drawing and losing a game against T2T_{2} are 12,16\frac{1}{2}, \frac{1}{6} and 13\frac{1}{3}, respectively. Each team gets 33 points for a win, 11 point for a draw and 00 point for a loss in a game. Let XX and YY denote the total points scored by teams T1T_{1} and T2T_{2}, respectively, after two games.
Question: P(X>Y)P(X>Y) is

Options

  1. A14\dfrac{1}{4}
  2. B512\dfrac{5}{12}
  3. C12\dfrac{1}{2}
  4. D712\dfrac{7}{12}

Correct answer

B. 512\dfrac{5}{12}

Step-by-step solution

P(X>Y)=P(T1win)P(T1win)+P(T1win)P(matchdraw)+P(matchdraw).P(T1win)P\left(X > Y\right) = P\left({T}_{1}win\right)P\left({T}_{1}win\right) + P\left({T}_{1}win\right)P\left(matchdraw\right) + P\left(matchdraw\right) . P\left({T}_{1}win\right)
=12×12+12×16+16×12=512= \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{1}{2} \times \dfrac{1}{6} + \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{5}{12}
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