(a) Two pupils are chosen at random from a group of 4 boys and 5

Explanation

(a) 4 boys, 5 girls

P(both are boys) = P(1st is a boy and 2nd is a boy)

= \(\frac{4}{9} \times \frac{3}{8}\)

= \(\frac{1}{6}\)

(b) If p is the probability of a standard transistors, then \(p = 0.8 = \frac{4}{5}\)

If q is the probability of a transistor below standard, \(q = 0.2 = \frac{1}{5}\)

For 6 transistors, we have

\((p + q)^{6} = p^{6} + 6p^{5}q + 15p^{4}q^{2} + 20p^{3}q^{3} + 15p^{2}q^{4} + 6pq^{5} + q^{6}\)

(i) p(exactly 2 standard transistors) = \(15p^{2}q^{4}\)

= \(15 \times (\frac{4}{5})^{2} \times (\frac{1}{5})^{4}\)

= \(\frac{48}{3125}\)

(ii) p(exactly 1 standard transistor) = \(6pq^{5}\)

= \(6 \times \frac{4}{5} \times (\frac{1}{5})^{5}\)

= \(\frac{24}{15625}\)

(iii) p(at least 2 standard transistors) = 1 - [p(0 standard transistor) + p(1 standard transistor)]

p(0 standard transistor) = \(q^{6}\)

= \((\frac{1}{5})^{6} = \frac{1}{15625}\)

p(at least 2 standard transistors) = \(1 - [\frac{1}{15625} + \frac{24}{15625}]\)

= \(\frac{15600}{15625} = \frac{624}{625}\)

(iv) p(at most 2 standard transistors) = p(0 standard transistor) + p(1 standard transistor) + p(2 standard transistors)

= \(\frac{1}{15625} + \frac{24}{15625} + \frac{48}{3125}\)

= \(\frac{265}{15625} = \frac{53}{3125}\)