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In a football game, one team either wins or there is a tie. This means there are three possible outcomes: a win, a loss or a draw. Therefore, the probability of a draw is \(\frac{1}{3}\).
The given answer is incorrect. There are three possibilities: a win, a loss or a draw. So, the possibility of a draw is \(\frac{1}{3}\).
\(P\) (games end in draw)
\(\Rightarrow\) Team \(P\) wins and \(Q\) wins
\(P\) (\(P\) wins) \(= \frac{1}{2}\)
\(P\) (\(Q\) wins) \(= \frac{1}{2}\)
Therefore, \(P\) (games ends in draw) \(= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)
Wrong