Question on: JAMB Mathematics - 1998
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?
A
\(\frac{2}{15}\)
B
\(\frac{7}{15}\)
C
\(\frac{11}{15}\)
D
\(\frac{13}{15}\)
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Correct Option: C
p(x) = \(\frac{2}{5}\) p(y) = \(\frac{1}{3}\)
p(x or y) = p(x ∪ y)
= p(x) + p(y)
= \(\frac{2}{5}\) + \(\frac{1}{3}\)
= \(\frac{11}{5}\)
p(x or y) = p(x ∪ y)
= p(x) + p(y)
= \(\frac{2}{5}\) + \(\frac{1}{3}\)
= \(\frac{11}{5}\)
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