| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Sampling distribution of range or maximum |
| Difficulty | Standard +0.8 This S2 question requires understanding of sampling distributions and systematic case analysis. Students must identify all possible samples of 3 balls (with replacement implied by 'large number'), calculate probabilities using the given ratio 2:3:4, determine the range for each case, and construct a complete probability distribution. The conceptual leap to enumerate cases systematically and the multi-step probability calculations make this moderately challenging, though it remains within standard S2 scope. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([A = \text{number on the ball}]\), \(P(A=1) = \frac{2}{9}\), \(P(A=2) = \frac{1}{3}\), \(P(A=5) = \frac{4}{9}\) | B1 | For writing or using the 3 correct probabilities |
| Possible samples with range of 4: \((1,1,5)\), \((1,2,5)\), \((1,5,5)\) | M1 | 1st M1: for identifying the 3 possible samples |
| \((1,1,5)\): \(\frac{2}{9} \times \frac{2}{9} \times \frac{4}{9} \times 3 = \frac{16}{243}\) or \((1,5,5)\): \(\frac{2}{9} \times \frac{4}{9} \times \frac{4}{9} \times 3 = \frac{32}{243}\) | M1 | 2nd M1: for \(p \times p \times q \times 3\) or \(p \times q \times q \times 3\) where \(p\) and \(q\) are probabilities with \((p+q) < 1\) |
| \((1,2,5)\): \(\frac{2}{9} \times \frac{1}{3} \times \frac{4}{9} \times 6 = \frac{16}{81}\) | M1 | 3rd M1: for \(p \times q \times r \times 6\) where \(p, q, r\) are probabilities with \((p+q+r) = 1\) |
| \(P(B=4) = \frac{16}{243} + \frac{32}{243} + \frac{16}{81} = \frac{32}{81}\) | A1 | For \(\frac{32}{81}\) or awrt 0.395 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(B=0) = \left(\frac{2}{9}\right)^3 + \left(\frac{1}{3}\right)^3 + \left(\frac{4}{9}\right)^3 = \frac{11}{81}\) | M1 | 1st M1: for \(p^3 + q^3 + r^3\) |
| \(P(B=1) = 3 \times \frac{2}{9} \times \left(\frac{1}{3}\right)^2 + 3 \times \frac{1}{3} \times \left(\frac{2}{9}\right)^2 = \frac{10}{81}\) or \(P(B=3) = 3 \times \frac{1}{3} \times \left(\frac{4}{9}\right)^2 + 3 \times \frac{4}{9} \times \left(\frac{1}{3}\right)^2 = \frac{28}{81}\) | M1 | 2nd M1: for \(3 \times p \times (q)^2 + 3 \times q \times (p)^2\) or \(3 \times q \times (r)^2 + 3 \times r \times (q)^2\) |
| \(1 - \frac{11}{81} - \frac{10}{81} - \frac{32}{81} = \frac{28}{81}\) or \(1 - \frac{11}{81} - \frac{28}{81} - \frac{32}{81} = \frac{10}{81}\) | M1 | 3rd M1: use of all probabilities of \(P(B=b)\) adding to 1 |
| Table with \(b\): 0, 1, 3, 4 | B1 | For ranges 0, 1, 3 and 4 with none omitted and no extras |
| \(P(B=b)\): \(\frac{11}{81}\), \(\frac{10}{81}\), \(\frac{28}{81}\), \(\frac{32}{81}\) | A1 | For a fully correct probability distribution |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[A = \text{number on the ball}]$, $P(A=1) = \frac{2}{9}$, $P(A=2) = \frac{1}{3}$, $P(A=5) = \frac{4}{9}$ | B1 | For writing or using the 3 correct probabilities |
| Possible samples with range of 4: $(1,1,5)$, $(1,2,5)$, $(1,5,5)$ | M1 | 1st M1: for identifying the 3 possible samples |
| $(1,1,5)$: $\frac{2}{9} \times \frac{2}{9} \times \frac{4}{9} \times 3 = \frac{16}{243}$ or $(1,5,5)$: $\frac{2}{9} \times \frac{4}{9} \times \frac{4}{9} \times 3 = \frac{32}{243}$ | M1 | 2nd M1: for $p \times p \times q \times 3$ or $p \times q \times q \times 3$ where $p$ and $q$ are probabilities with $(p+q) < 1$ |
| $(1,2,5)$: $\frac{2}{9} \times \frac{1}{3} \times \frac{4}{9} \times 6 = \frac{16}{81}$ | M1 | 3rd M1: for $p \times q \times r \times 6$ where $p, q, r$ are probabilities with $(p+q+r) = 1$ |
| $P(B=4) = \frac{16}{243} + \frac{32}{243} + \frac{16}{81} = \frac{32}{81}$ | A1 | For $\frac{32}{81}$ or awrt 0.395 |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B=0) = \left(\frac{2}{9}\right)^3 + \left(\frac{1}{3}\right)^3 + \left(\frac{4}{9}\right)^3 = \frac{11}{81}$ | M1 | 1st M1: for $p^3 + q^3 + r^3$ |
| $P(B=1) = 3 \times \frac{2}{9} \times \left(\frac{1}{3}\right)^2 + 3 \times \frac{1}{3} \times \left(\frac{2}{9}\right)^2 = \frac{10}{81}$ or $P(B=3) = 3 \times \frac{1}{3} \times \left(\frac{4}{9}\right)^2 + 3 \times \frac{4}{9} \times \left(\frac{1}{3}\right)^2 = \frac{28}{81}$ | M1 | 2nd M1: for $3 \times p \times (q)^2 + 3 \times q \times (p)^2$ or $3 \times q \times (r)^2 + 3 \times r \times (q)^2$ |
| $1 - \frac{11}{81} - \frac{10}{81} - \frac{32}{81} = \frac{28}{81}$ or $1 - \frac{11}{81} - \frac{28}{81} - \frac{32}{81} = \frac{10}{81}$ | M1 | 3rd M1: use of all probabilities of $P(B=b)$ adding to 1 |
| Table with $b$: 0, 1, 3, 4 | B1 | For ranges 0, 1, 3 and 4 with none omitted and no extras |
| $P(B=b)$: $\frac{11}{81}$, $\frac{10}{81}$, $\frac{28}{81}$, $\frac{32}{81}$ | A1 | For a fully correct probability distribution |
---\begin{enumerate}
\item A bag contains a large number of balls, each with one of the numbers 1,2 or 5 written on it in the ratio $2 : 3 : 4$ respectively.
\end{enumerate}
A random sample of 3 balls is taken from the bag.\\
The random variable $B$ represents the range of the numbers written on the balls in the sample.\\
(i) Find $\mathrm { P } ( B = 4 )$\\
(ii) Find the sampling distribution of $B$.
\hfill \mbox{\textit{Edexcel S2 2021 Q4 [10]}}