(a) p(man wins) = p = 0.8 ; p(man loses) = q = 0.2.
4 different races : \((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2} q^{2} + 4pq^{3} + q^{4}\)
(i) p(man wins all) = \(p^{4} = (0.8)^{4} = 0.4096\)
(ii) p(man wins no race) = \(q^{4} = (0.2)^{4} = 0.0016\)
(iii) p(man wins at most 3) = 1 - p(man wins all)
= 1 - 0.4096 = 0.5904.
(b)(i) 5 girls, 10 boys ; Total = 15.
Ways of selecting 3 boys = \(^{10}C_{3} \times ^{5}C_{2}\)
= \(120 times 10\)
= 1200
Without restriction, selection = \(^{15}C_{5}\)
= \(\frac{15!}{10! 5!}\)
= 3003
p(selecting 3 boys) = \(\frac{1200}{3003} = 0.3996\)
(ii) Ways of selecting no girls = \(^{10}C_{5} \times ^{5}C_{0}\)
= 252
p(at least one girl) = 1 - p(no girl)
= \(1 - \frac{252}{3003}\)
= \(2751}{3003} = 0.9161\)