A basket contains 9 apples, 8 bananas and 7 oranges. A fruit is picked from the basket, find the probability that it is neither an apple nor an orange.
\(\frac{3}{8}\)
\(\frac{1}{3}\)
\(\frac{7}{24}\)
\(\frac{2}{3}\)
Correct answer is B
n(apples) = 9
n(bananas) = 8
n(oranges) = 7
n(\(\varepsilon\)) = 24
Hence Prob(not apple, nor orange) = Prob(banana) = \(\frac{8}{24}\) = \(\frac{1}{3}\)