| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Expected profit or cost problem |
| Difficulty | Moderate -0.3 This is a standard S2 probability distribution question with routine calculations: basic combinatorics for probabilities, constructing a probability distribution table, calculating expectation using E(X) = Σxp(x), and applying linear transformation properties E(aX+b) and SD(aX+b). All techniques are textbook exercises with no novel insight required, though the multi-part structure and careful arithmetic make it slightly below average difficulty rather than trivial. |
| 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{GG or YY or RR}) = \frac{2}{10}\times\frac{1}{9} + \frac{3}{10}\times\frac{2}{9} + \frac{4}{10}\times\frac{3}{9}\) | M1 | |
| \(= \frac{2}{9}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(B\bar{B}\ \text{or}\ \bar{B}B) = \frac{1}{10}\times\frac{9}{9} + \frac{9}{10}\times\frac{1}{9}\) | M1 | \(\frac{1}{10} + \frac{9}{10}\times\frac{1}{9}\) |
| \(= \frac{1}{5}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Table: \(x\): Same \(= 135\), 1 Blue \(= 145\), Neither \(= -45\) | B1 | |
| \(P(X=x)\): \(\frac{2}{9},\ \frac{1}{5},\ \frac{26}{45}\) | B1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = 135\times\frac{2}{9} + 145\times\frac{1}{5} + (-45)\times\frac{26}{45}\) | M1 | Multiply two rows of their table from (b)(i) |
| \(= 29 + 30 - 26 = 33\) pence | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y) = 104 - 3E(X)\) | M1 | |
| \(= 104 - 3\times 33 = 5\) pence | A1 | |
| \(\therefore\) Joanne would expect to win £5 | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^2) = 9425\) | B1 | \((4205 + 4050 + 1170)\) |
| \(\text{Var}(X) = 9425 - 33^2 = \mathbf{8336}\) | \(sd(X) = 91.30\) | |
| \(\text{Var}(Y) = 9 \times \text{Var}(X)\) | B1 | \(9\times(\text{their Var}(X) > 0)\) |
| \(= 9 \times 8336 = 75024\) | M1 | or \(3\times(\text{their } sd(X))\) |
| \(\Rightarrow\) standard deviation \((Y) = 274\) pence | A1 | 4 |
| Total | 15 |
# Question 5(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{GG or YY or RR}) = \frac{2}{10}\times\frac{1}{9} + \frac{3}{10}\times\frac{2}{9} + \frac{4}{10}\times\frac{3}{9}$ | M1 | |
| $= \frac{2}{9}$ | A1 | 2 | (AG) |
---
# Question 5(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(B\bar{B}\ \text{or}\ \bar{B}B) = \frac{1}{10}\times\frac{9}{9} + \frac{9}{10}\times\frac{1}{9}$ | M1 | $\frac{1}{10} + \frac{9}{10}\times\frac{1}{9}$ |
| $= \frac{1}{5}$ | A1 | 2 | (AG) |
---
# Question 5(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Table: $x$: Same $= 135$, 1 Blue $= 145$, Neither $= -45$ | B1 | |
| $P(X=x)$: $\frac{2}{9},\ \frac{1}{5},\ \frac{26}{45}$ | B1 | 2 | |
---
# Question 5(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = 135\times\frac{2}{9} + 145\times\frac{1}{5} + (-45)\times\frac{26}{45}$ | M1 | Multiply two rows of their table from (b)(i) |
| $= 29 + 30 - 26 = 33$ pence | A1 | 2 | AG |
---
# Question 5(c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y) = 104 - 3E(X)$ | M1 | |
| $= 104 - 3\times 33 = 5$ pence | A1 | |
| $\therefore$ Joanne would expect to win £5 | A1 | 3 | OE (eg 500p) |
---
# Question 5(c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^2) = 9425$ | B1 | $(4205 + 4050 + 1170)$ |
| $\text{Var}(X) = 9425 - 33^2 = \mathbf{8336}$ | | $sd(X) = 91.30$ |
| $\text{Var}(Y) = 9 \times \text{Var}(X)$ | B1 | $9\times(\text{their Var}(X) > 0)$ |
| $= 9 \times 8336 = 75024$ | M1 | **or** $3\times(\text{their } sd(X))$ |
| $\Rightarrow$ standard deviation $(Y) = 274$ pence | A1 | 4 | 273.9p or £2.74 |
| **Total** | **15** | |
---5 Joanne has 10 identically-shaped discs, of which 1 is blue, 2 are green, 3 are yellow and 4 are red. She places the 10 discs in a bag and asks her friend David to play a game by selecting, at random and without replacement, two discs from the bag.
\begin{enumerate}[label=(\alph*)]
\item Show that:
\begin{enumerate}[label=(\roman*)]
\item the probability that the two discs selected are the same colour is $\frac { 2 } { 9 }$;
\item the probability that exactly one of the two discs selected is blue is $\frac { 1 } { 5 }$.
\end{enumerate}\item Using the discs, Joanne plays the game with David, under the following conditions:
If the two discs selected by David are the same colour, she will pay him 135p. If exactly one of the two discs selected by David is blue, she will pay him 145p. Otherwise David will pay Joanne 45p.
\begin{enumerate}[label=(\roman*)]
\item When a game is played, $X$ is the amount, in pence, won by David. Construct the probability distribution for $X$, in the form of a table.
\item Show that $\mathrm { E } ( X ) = 33$.
\end{enumerate}\item Joanne modifies the game so that the amount per game, $Y$ pence, that she wins may be modelled by
$$Y = 104 - 3 X$$
\begin{enumerate}[label=(\roman*)]
\item Determine how much Joanne would expect to win if the game is played 100 times.
\item Calculate the standard deviation of $Y$, giving your answer to the nearest 1 p .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2009 Q5 [15]}}