There are 8 boys and 6 girls in a class. If two students are selected at random from the class, find the probability that they are of
(a) the same sex ;
(b) different sex.
\(p(boys) = \frac{8}{14} = \frac{4}{7}\)
\(p(girls) = \frac{6}{14} = \frac{3}{7}\)
(a) p(both are of the same sex) = p(both are boys or both are girls)
= \(\frac{8}{14} \times \frac{7}{13} + \frac{6}{14} \times \frac{5}{13}\)
= \(\frac{86}{182}\)
= \(\frac{43}{91}\)
(b) p(they are of different sex) = p(first is a boy and second is a girl or first is a girl and second is a boy)
= \(\frac{8}{14} \times \frac{6}{13} + \frac{6}{14} \times \frac{8}{13}\)
= \(\frac{96}{182}\)
= \(\frac{48}{91}\)
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