Three school prefects are to be chosen from four girls and five boys. What is the probability that :
(a) only boys will be chosen ;
(b) more girls than boys will be chosen ?
Explanation
4 girls and 5 boys
(a) p(only boys) = (frac{^{5}C_{3}}{^{9}C_{3}})
= (frac{5!}{2!3!} times frac{3!6!}{9!})
= (frac{10 times 1}{84})
= (frac{5}{42})
(b) p(more girls than boys) = p(3 girls and no boys) or p(2 girls and 1 boy)
Number of ways of selecting = (^{4}C_{3} times ^{5}C_{0} + ^{4}C_{2} times ^{5}C_{1})
= (4 times 1 + 6 times 5)
= 34 ways.
Without restrictions, selection can be made in (^{9}C_{3}) ways.
= (frac{9!}{6! 3!})
= 84 ways.
(therefore) p(more girls than boys) = (frac{34}{84})
= (frac{17}{42}).