JAMB - Mathematics (1998 - No. 36)
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?
\(\frac{2}{15}\)
\(\frac{7}{15}\)
\(\frac{11}{15}\)
\(\frac{13}{15}\)
Explanation
To find the probability that either team X or team Y wins the gold, we can use the formula for the probability of the union of two events: P(x or y) = p(x) + p(y) - p(x n y)
But p(x n y ) = 0
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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