In a recent zonal championship games involving 10 teams, teams X and Y were given probabilities \(\frac{2}{5}\) and \(\frac{1}{3}\) respectively of winning the gold in the football event. What is the probability that either team will win the gold?

\(\frac{2}{15}\)

\(\frac{7}{15}\)

\(\frac{11}{15}\)

\(\frac{13}{15}\)

Correct answer is C

p(x) = \(\frac{2}{5}\) p(y) = \(\frac{1}{3}\)

p(x or y) = p(x ∪ y)

= p(x) + p(y)

= \(\frac{2}{5}\) + \(\frac{1}{3}\)

= \(\frac{11}{5}\)

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