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JEE Advanced · Mathematics · 32. Probability

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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.


Question:

\(P(X>Y)\) is

  1. A 14
  2. B 512
  3. C 12
  4. D 712
Verified Solution

Answer & Solution

Correct Answer

(B) 512

Step-by-step Solution

Detailed explanation

PX>Y=PT1 win PT1win+PT1win Pmatch draw+Pmatch draw. PT1win
=12×12+12×16+16×12=512
From JEE Advanced
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