| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Substitution into binomial expansion |
| Difficulty | Standard +0.8 This question requires identifying that 'Alison ahead' means winning at least 3 of 5 points (binomial probabilities for X=3,4,5), computing these using C(5,r)p^r(1-p)^(5-r), then algebraically simplifying the sum to match the given form p^3(6p^2-15p+10). Part (b) is routine substitution. The algebraic manipulation to reach the exact form shown requires careful expansion and factoring, making this harder than standard binomial probability questions but not exceptionally difficult for A-level. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks |
|---|---|
| (a) \(X \sim B(5, p)\) | B1 |
| \[P(X \ge 3) = \binom{5}{3}p^3(1-p)^2 + \binom{5}{4}p^4(1-p) + \binom{5}{5}p^5(1-p)^0\] | M1 A1 A1 |
| \[= p^5 + 5p^4 - 5p^3 + 10p^3(1-2p + p^2)\] | A1 |
| \[= 6p^5 - 15p^4 + 10p^3 = p^3(6p^2 - 15p + 10)\] | M1 A1 |
| (b) Put \(p = 0.6\) to get \(P(X \ge 3) = 0.683\) | M1 A1 |
(a) $X \sim B(5, p)$ | B1 |
$$P(X \ge 3) = \binom{5}{3}p^3(1-p)^2 + \binom{5}{4}p^4(1-p) + \binom{5}{5}p^5(1-p)^0$$ | M1 A1 A1 |
$$= p^5 + 5p^4 - 5p^3 + 10p^3(1-2p + p^2)$$ | A1 |
$$= 6p^5 - 15p^4 + 10p^3 = p^3(6p^2 - 15p + 10)$$ | M1 A1 |
(b) Put $p = 0.6$ to get $P(X \ge 3) = 0.683$ | M1 A1 |
**Total: 9 marks**4. Alison and Gemma play table tennis. Alison starts by serving for the first five points. The probability that she wins a point when serving is $p$.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that Alison is ahead at the end of her five serves is given by
$$p ^ { 3 } \left( 6 p ^ { 2 } - 15 p + 10 \right) .$$
\item Evaluate this probability when $p = 0.6$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q4 [9]}}