(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.
Show Answer Show Explanation 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)