(a) Simplify (^{n + 1}C_{3} – ^{n – 1}C_{3})
(b) A fair die is thrown five times. Calculate, correct to three decimal places, the probability of obtaining (i) at most two sixes ; (ii) exactly three sixes.
Explanation
(a) (^{n + 1}C_{3} – ^{n – 1}C_{3})
= (frac{(n + 1)!}{3! (n + 1 – 3)!} – frac{(n – 1)!}{3! (n – 1 – 3)!})
= (frac{(n + 1)!}{3! (n – 2)!} – frac{(n – 1)!}{3! (n – 4)!})
= (frac{(n + 1)(n)(n – 1)(n – 2)!}{3! (n – 2)!} – frac{(n – 1)(n – 2)(n – 3)(n – 4)!}{3! (n – 4)!})
= (frac{(n + 1)(n)(n – 1)}{6} – frac{(n – 1)(n – 2)(n – 3)}{6})
= (frac{(n – 1)[(n + 1)(n) – (n – 2)(n – 3)]}{6})
= (frac{(n – 1)[n^{2} + n – (n^{2} – 5n + 6)}{6})
= (frac{(n – 1)[6n – 6]}{6})
= (frac{6(n – 1)^{2}}{6})
= ((n – 1)^{2})
(b) p(six) = p = (frac{1}{6}), p(not six) = q = (frac{5}{6}).
((p + q)^{5} = p^{5} + 5p^{4} q + 10p^{3} q^{2} + 10p^{2} q^{3} + 5pq^{4} + q^{5})
(i) p( at most 2 sixes) = (q^{5} + 5pq^{4} + 10p^{2} q^{3})
= ((frac{5}{6})^{5} + 5(frac{1}{6})(frac{5}{6})^{4} + 10(frac{1}{6})^{2} (frac{5}{6})^{3})
= (frac{3125}{7776} + frac{3125}{7776} + frac{1250}{7776})
= (frac{7500}{7776} = 0.9645 approxeq 0.965) (3 d.p)
(ii) p(exactly 3 sixes) = (10p^{3} q^{2})
= (10(frac{1}{6})^{3} (frac{5}{6})^{2})
= (frac{250}{7776})
= (0.032) (3 d.p)