| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Probability distribution from tree |
| Difficulty | Standard +0.8 This S2 question requires constructing a tree diagram with dependent probabilities, forming an algebraic equation from a given probability condition, solving for an unknown parameter, then computing a complete probability distribution. It demands careful case analysis across multiple branches and systematic probability calculations, going beyond routine tree diagram exercises. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\left(2\times\frac{a}{a+7}\times\frac{1}{4}\times\frac{3}{4}\right) + \left(\frac{a}{a+7}\times\frac{1}{4}\times\frac{1}{4}\right) = \frac{63}{256}\) or \(\frac{a}{a+7}\times\left(1-\left(\frac{3}{4}\right)^2\right) = \frac{63}{256}\) | M1 | For use of \(\frac{a}{a+7}\) in an equation |
| \(\frac{a}{a+7}\times\left[\left(\frac{1}{4}\right)^2 + 2\times\frac{1}{4}\times\frac{3}{4}\right] = \frac{63}{256}\) | M1 | Setting up a correct equation to find the value of \(a\) |
| \(\frac{a}{7+a} = \frac{63}{256} \div \frac{7}{16}\) or \(\frac{a}{7+a} = \frac{9}{16}\) \(\therefore a = 9\) | A1* | \(a = 9\) with at least one further correct line of working |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Range \((R)\): 0, 5, 10 (and 15) | B1 | For the 3 ranges 0, 5 and 10 and no extra incorrect ones, or all extras have probability 0 |
| \(P(20) = \frac{9}{16}\), \(P(5) = \frac{5}{16}\), \(P(10) = \frac{2}{16}\) | B1 | For the correct 3 probabilities; \(\frac{2}{16}\) and/or \(\frac{5}{16}\) may be implied by correct answer for \(P(R=0)\) or \(P(R=5)\); \(\frac{9}{16}\) may be implied by correct answer for \(P(R=10)\) |
| \(P(R=0) = \frac{5}{16}\times\frac{1}{4}\times\frac{1}{4} + \frac{2}{16}\times\frac{3}{4}\times\frac{3}{4}\) | M1 | Correct method for one probability |
| \(P(R=5) = 2\times\frac{5}{16}\times\frac{1}{4}\times\frac{3}{4} + 2\times\frac{2}{16}\times\frac{1}{4}\times\frac{3}{4} + \frac{2}{16}\times\frac{1}{4}\times\frac{1}{4}\) | M1 | Correct method for two probabilities |
| \(P(R=10) = \frac{9}{16}\times\frac{3}{4}\times\frac{3}{4}\) | M1 | Correct method for all 3 probabilities |
| \(R\): 0, 5, 10, 15 with \(r\): \(\frac{23}{256}\), \(\frac{89}{256}\), \(\frac{81}{256}\), \(\frac{63}{256}\) | A1cao | Correct ranges with correct associated probabilities; allow decimals: 0.090, 0.348, 0.316, 0.246 |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left(2\times\frac{a}{a+7}\times\frac{1}{4}\times\frac{3}{4}\right) + \left(\frac{a}{a+7}\times\frac{1}{4}\times\frac{1}{4}\right) = \frac{63}{256}$ or $\frac{a}{a+7}\times\left(1-\left(\frac{3}{4}\right)^2\right) = \frac{63}{256}$ | M1 | For use of $\frac{a}{a+7}$ in an equation |
| $\frac{a}{a+7}\times\left[\left(\frac{1}{4}\right)^2 + 2\times\frac{1}{4}\times\frac{3}{4}\right] = \frac{63}{256}$ | M1 | Setting up a correct equation to find the value of $a$ |
| $\frac{a}{7+a} = \frac{63}{256} \div \frac{7}{16}$ or $\frac{a}{7+a} = \frac{9}{16}$ $\therefore a = 9$ | A1* | $a = 9$ with at least one further correct line of working |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Range $(R)$: 0, 5, 10 (and 15) | B1 | For the 3 ranges 0, 5 and 10 and no extra incorrect ones, or all extras have probability 0 |
| $P(20) = \frac{9}{16}$, $P(5) = \frac{5}{16}$, $P(10) = \frac{2}{16}$ | B1 | For the correct 3 probabilities; $\frac{2}{16}$ and/or $\frac{5}{16}$ may be implied by correct answer for $P(R=0)$ or $P(R=5)$; $\frac{9}{16}$ may be implied by correct answer for $P(R=10)$ |
| $P(R=0) = \frac{5}{16}\times\frac{1}{4}\times\frac{1}{4} + \frac{2}{16}\times\frac{3}{4}\times\frac{3}{4}$ | M1 | Correct method for one probability |
| $P(R=5) = 2\times\frac{5}{16}\times\frac{1}{4}\times\frac{3}{4} + 2\times\frac{2}{16}\times\frac{1}{4}\times\frac{3}{4} + \frac{2}{16}\times\frac{1}{4}\times\frac{1}{4}$ | M1 | Correct method for two probabilities |
| $P(R=10) = \frac{9}{16}\times\frac{3}{4}\times\frac{3}{4}$ | M1 | Correct method for all 3 probabilities |
| $R$: 0, 5, 10, 15 with $r$: $\frac{23}{256}$, $\frac{89}{256}$, $\frac{81}{256}$, $\frac{63}{256}$ | A1cao | Correct ranges with correct associated probabilities; allow decimals: 0.090, 0.348, 0.316, 0.246 |
---\begin{enumerate}
\item A bag contains a large number of counters with one of the numbers 5 , 10 or 20 written on each of them in the ratio $5 : 2 : a$
\end{enumerate}
A jar contains a large number of counters with one of the numbers 5 or 10 written on each of them in the ratio $1 : 3$
One counter is selected at random from the bag and then two counters are selected at random from the jar.\\
The random variable $R$ represents the range of the numbers on the 3 counters.\\
Given that $\mathrm { P } ( R = 15 ) = \frac { 63 } { 256 }$\\
(a) by forming and solving an equation in $a$, show that $a = 9$\\
(b) find the sampling distribution of $R$
\hfill \mbox{\textit{Edexcel S2 2022 Q6 [9]}}