(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...
Question 1 Report
(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.
Answer Details
(a) Write each combination out:
\[{}^{n+1}C_3 = \frac{(n+1)n(n-1)}{6},\qquad {}^{n-1}C_3 = \frac{(n-1)(n-2)(n-3)}{6}\] \[{}^{n+1}C_3 - {}^{n-1}C_3 = \frac{n-1}{6}\Big[(n+1)n - (n-2)(n-3)\Big]\] \[= \frac{n-1}{6}\Big[(n^2 + n) - (n^2 - 5n + 6)\Big] = \frac{n-1}{6}(6n - 6) = (n - 1)^2\]So the expression simplifies to \((n - 1)^2\).
(b) A six has \(p = \dfrac16,\ q = \dfrac56\), with \(n = 5\) throws (binomial).
(i) At most two sixes \(= P(0) + P(1) + P(2)\):
\[P(0) = \left(\tfrac56\right)^5 = \tfrac{3125}{7776},\quad P(1) = \binom{5}{1}\tfrac16\left(\tfrac56\right)^4 = \tfrac{3125}{7776},\quad P(2) = \binom{5}{2}\left(\tfrac16\right)^2\left(\tfrac56\right)^3 = \tfrac{1250}{7776}\] \[P(\le 2) = \frac{3125 + 3125 + 1250}{7776} = \frac{7500}{7776} \approx 0.965\](ii) Exactly three sixes:
\[P(3) = \binom{5}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^2 = 10\cdot\frac{1}{216}\cdot\frac{25}{36} = \frac{250}{7776} \approx 0.032\]Download The App On Google Playstore
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