| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Nested binomial expected count |
| Difficulty | Standard +0.8 This is a multi-layered problem requiring understanding of sampling without replacement, binomial distribution applied to a derived probability, and systematic enumeration of sample spaces. Part (a) requires calculating P(9 in sample) using combinations, then applying binomial mean/variance formulas. Parts (b)-(c) demand careful enumeration of all possible samples and their medians—non-trivial combinatorial reasoning beyond standard textbook exercises. The nested structure (binomial on top of hypergeometric sampling) and the median distribution calculation elevate this above typical S2 questions. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(Y \sim B(20, p)\) \(p = P(\text{sample contains counter with a 9 on it})\) | M1, A1 | |
| \(p = \left(1 - \frac{9}{10} × \frac{8}{9} × \frac{7}{8}\right)\) oe or \(\left(\frac{9}{10} × \frac{8}{9} × \frac{7}{8}\right)\) oe or \(\frac{6}{10} × \frac{5}{9} × \frac{1}{8} × 3 + \frac{6}{10} × \frac{3}{9} × \frac{1}{8} × 6 + \frac{3}{10} × \frac{2}{9} × \frac{1}{8} × 3\) oe \(= \left[\frac{3}{10}\right]\) | 1st M1 For all methods condone missing ×3 and /or ×6. Allow \(^1C_3, ^2C_3\) oe. Condone with replacement - condone missing ×3 and or ×6. | |
| (i) \(E(Y) = 20 × "a"_1 = [6]\) | B1 | B1 for \(20 ×\) probability \(–\) no need to calculate |
| (ii) \(\text{Var}(Y) = 20 × "a_1" × (1 - "a_1") = 4.2\) | M1, A1 | 2nd M1 Use of \(np(1 - p)\) or \(np\left(1 - \frac{np}{20}\right)\). 2nd A1 variance = 4.2 |
| (b) \((7,7,7)\) | B2 | B1B1 all 3 correct (with none incorrect – ignore arrangements of the correct numbers) (B1B0 any one correct and no incorrect or 2 or 3 correct and only one incorrect)These can be awarded in part (c) provided that they are clearly identified as having a median of 7 More than one incorrect is B0B0 |
| \((7,7,8), [(7,8,7), (8,7,7)]\) | ||
| \((7,7,9), [(7,9,7), (9,7,7)]\) | ||
| (c) | m | 7 |
| \(P(M = m)\) | \(\frac{6}{10} × \frac{5}{9} × \frac{4}{8} + 3 × \frac{6}{10} × \frac{5}{9} × \frac{1}{8} + 3 × \frac{6}{10} × \frac{3}{9} × \frac{1}{8} × 3\) oe | \(1 - P(M = 1)\) |
| \(= \frac{2}{3}\) | \(= \frac{1}{3}\) | |
| B1, M1 | B1 for identifying that the only possible medians are 7 and 8. Allow 9 if it has a probability of 0. 1st M1 correct expression for P(M = 7) Implied by 2/3 or P(M = 8) Implied by 1/3 P(M = 8) = \(\left(\frac{3}{10}\right)^3 + 3 × \frac{6}{10} × \left(\frac{3}{10}\right)^2 + 3 × \left(\frac{3}{10}\right)^2 × \frac{1}{10} - 0.648\) or P(M = 8) = \(\left(\frac{3}{10}\right)^3 + 3 × \frac{6}{10} × \left(\frac{3}{10}\right)^2 + 3 × \left(\frac{3}{10}\right)^2 × \frac{1}{10} + 6 × \frac{6}{10} × \frac{3}{10} × \frac{1}{10} = 0.324\) | |
| A1, A1 | ||
| 2nd M1 Total of the 2 probabilities for 7 and 8 = 1 or a correct expression without replacement for both P(M = 7) and P(M = 8) condone with replacement. 1st A1 P(M = 7) = \(\frac{2}{3}\) oe 2nd A1 P(M = 8) = \(\frac{1}{3}\) oe |
**(a)** $Y \sim B(20, p)$ $p = P(\text{sample contains counter with a 9 on it})$ | M1, A1 |
$p = \left(1 - \frac{9}{10} × \frac{8}{9} × \frac{7}{8}\right)$ oe or $\left(\frac{9}{10} × \frac{8}{9} × \frac{7}{8}\right)$ oe or $\frac{6}{10} × \frac{5}{9} × \frac{1}{8} × 3 + \frac{6}{10} × \frac{3}{9} × \frac{1}{8} × 6 + \frac{3}{10} × \frac{2}{9} × \frac{1}{8} × 3$ oe $= \left[\frac{3}{10}\right]$ | | 1st M1 For all methods condone missing ×3 and /or ×6. Allow $^1C_3, ^2C_3$ oe. Condone with replacement - condone missing ×3 and or ×6.
**(i)** $E(Y) = 20 × "a"_1 = [6]$ | B1 | B1 for $20 ×$ probability $–$ no need to calculate
**(ii)** $\text{Var}(Y) = 20 × "a_1" × (1 - "a_1") = 4.2$ | M1, A1 | 2nd M1 Use of $np(1 - p)$ or $np\left(1 - \frac{np}{20}\right)$. 2nd A1 variance = 4.2
**(b)** $(7,7,7)$ | B2 | B1B1 all 3 correct (with none incorrect – ignore arrangements of the correct numbers) (B1B0 any one correct and no incorrect or 2 or 3 correct and only one incorrect)These can be awarded in part (c) provided that they are clearly identified as having a median of 7 More than one incorrect is B0B0
$(7,7,8), [(7,8,7), (8,7,7)]$ | |
$(7,7,9), [(7,9,7), (9,7,7)]$ | |
**(c)** | m | 7 | 8 |
|---|---|---|---|
| $P(M = m)$ | $\frac{6}{10} × \frac{5}{9} × \frac{4}{8} + 3 × \frac{6}{10} × \frac{5}{9} × \frac{1}{8} + 3 × \frac{6}{10} × \frac{3}{9} × \frac{1}{8} × 3$ oe | $1 - P(M = 1)$ |
| | $= \frac{2}{3}$ | $= \frac{1}{3}$ |
| B1, M1 | B1 for identifying that the only possible medians are 7 and 8. Allow 9 if it has a probability of 0. 1st M1 correct expression for P(M = 7) Implied by 2/3 or P(M = 8) Implied by 1/3 P(M = 8) = $\left(\frac{3}{10}\right)^3 + 3 × \frac{6}{10} × \left(\frac{3}{10}\right)^2 + 3 × \left(\frac{3}{10}\right)^2 × \frac{1}{10} - 0.648$ or P(M = 8) = $\left(\frac{3}{10}\right)^3 + 3 × \frac{6}{10} × \left(\frac{3}{10}\right)^2 + 3 × \left(\frac{3}{10}\right)^2 × \frac{1}{10} + 6 × \frac{6}{10} × \frac{3}{10} × \frac{1}{10} = 0.324$ | |
| | | | | A1, A1 |
| | 2nd M1 Total of the 2 probabilities for 7 and 8 = 1 or a correct expression without replacement for both P(M = 7) and P(M = 8) condone with replacement. 1st A1 P(M = 7) = $\frac{2}{3}$ oe 2nd A1 P(M = 8) = $\frac{1}{3}$ oe |
**[Total 12]**\begin{enumerate}
\item A bag contains 10 counters each with exactly one number written on it.
\end{enumerate}
There are 6 counters with the number 7 on them\\
There are 3 counters with the number 8 on them\\
There is 1 counter with the number 9 on it\\
A random sample of 3 counters is taken from the bag (without replacement).\\
These counters are then put back in the bag.\\
This process is then repeated until 20 samples have been taken.\\
The random variable $Y$ represents the number of these 20 samples that contain the counter with the number 9 on it.\\
(a) (i) Find the mean of $Y$\\
(ii) Find the variance of $Y$
A random sample of 3 counters is chosen from the bag (without replacement).\\
(b) List all possible samples where the median of the numbers on the 3 counters is 7\\
(c) Find the sampling distribution of the median of the numbers on the 3 counters.
\hfill \mbox{\textit{Edexcel S2 2022 Q7 [12]}}