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?
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Correct Answer: Option C
Explanation:
p(x) = \(\frac{2}{5}\) p(y) = \(\frac{1}{3}\)
p(x or y) = p(x ∪ y)
= p(x) + p(y)
= \(\frac{2}{5}\) + \(\frac{1}{3}\)
= \(\frac{11}{5}\)
p(x) = \(\frac{2}{5}\) p(y) = \(\frac{1}{3}\)
p(x or y) = p(x ∪ y)
= p(x) + p(y)
= \(\frac{2}{5}\) + \(\frac{1}{3}\)
= \(\frac{11}{5}\)