| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | Calculate expectation and variance |
| Difficulty | Moderate -0.8 This is a straightforward hypergeometric distribution question requiring basic probability axioms, expectation by symmetry, variance calculation from a given distribution, and simple expected value computation. All parts follow standard S2 procedures with no novel problem-solving required, making it easier than average A-level material. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(\boldsymbol { x }\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | \(\frac { 1 } { 252 }\) | \(\frac { 25 } { 252 }\) | \(\frac { 100 } { 252 }\) | \(a\) | \(\frac { 25 } { 252 }\) | \(\frac { 1 } { 252 }\) |
| \(\boldsymbol { y }\) | 0 | 1 | 2 |
| \(\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )\) | \(\frac { 2 } { 9 }\) | \(\frac { 5 } { 9 }\) | \(\frac { 2 } { 9 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = \frac{25}{63}\) (OE) | B1 | \(\frac{100}{252}\) or \(\frac{50}{126}\) or \(0.397\) |
| Answer | Marks |
|---|---|
| \(E(X) = 2.5\) (symmetry) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = \left(1 \times \frac{25}{252}\right) + \left(4 \times \frac{25}{63}\right) + \left(9 \times \frac{25}{63}\right) + \left(16 \times \frac{25}{252}\right) + \left(25 \times \frac{1}{252}\right)\) | M1 | \(\sum x^2 \times p\) attempted |
| \(E(X^2) = \frac{125}{18}\) | A1 | \(\left(6\frac{17}{18}\) or \(6.94\right)\) |
| \(\text{Var}(X) = \frac{125}{18} - \frac{25}{4} = \frac{25}{36}\) | m1, A1 | dep \(\sum x^2 \times p\) used; \([\text{their } E(X^2) - (\text{their } E(X))^2]\); Var \(> 0\) |
| \(\text{sd}(X) = \frac{5}{6}\) | A1ft | \(0.83\dot{3}\) \(\left(\sqrt{\text{their Var}(X)}\right)\) (dep m1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(\text{Pay}) = \frac{4}{9} \times 90\) pence \(= 40\) pence \(\Rightarrow\) Joanne expected to make a loss (loss of 10p per game) | M1, A1 | Alternative: \(\frac{5}{9} > \frac{2}{9} + \frac{2}{9} \Rightarrow\) loss (for B1) then M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(\text{Loss}) = 100 \times 10\) pence \(= £10\) | B1ft | \(100 \times (\text{their loss/game})\) |
## Question 6(a)(i):
| $a = \frac{25}{63}$ (OE) | B1 | $\frac{100}{252}$ or $\frac{50}{126}$ or $0.397$ |
## Question 6(a)(ii):
| $E(X) = 2.5$ (symmetry) | B1 | |
## Question 6(a)(iii):
| $E(X^2) = \left(1 \times \frac{25}{252}\right) + \left(4 \times \frac{25}{63}\right) + \left(9 \times \frac{25}{63}\right) + \left(16 \times \frac{25}{252}\right) + \left(25 \times \frac{1}{252}\right)$ | M1 | $\sum x^2 \times p$ attempted |
| $E(X^2) = \frac{125}{18}$ | A1 | $\left(6\frac{17}{18}$ or $6.94\right)$ |
| $\text{Var}(X) = \frac{125}{18} - \frac{25}{4} = \frac{25}{36}$ | m1, A1 | dep $\sum x^2 \times p$ used; $[\text{their } E(X^2) - (\text{their } E(X))^2]$; Var $> 0$ |
| $\text{sd}(X) = \frac{5}{6}$ | A1ft | $0.83\dot{3}$ $\left(\sqrt{\text{their Var}(X)}\right)$ (dep **m1**) |
## Question 6(b)(i):
| $E(\text{Pay}) = \frac{4}{9} \times 90$ pence $= 40$ pence $\Rightarrow$ Joanne expected to make a loss (loss of 10p per game) | M1, A1 | Alternative: $\frac{5}{9} > \frac{2}{9} + \frac{2}{9} \Rightarrow$ loss (for B1) then M1A1 |
## Question 6(b)(ii):
| $E(\text{Loss}) = 100 \times 10$ pence $= £10$ | B1ft | $100 \times (\text{their loss/game})$ |
---6
\begin{enumerate}[label=(\alph*)]
\item Ali has a bag of 10 balls, of which 5 are red and 5 are blue. He asks Ben to select 5 of these balls from the bag at random.
The probability distribution of $X$, the number of red balls that Ben selects, is given in Table 1.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 1}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & $\frac { 1 } { 252 }$ & $\frac { 25 } { 252 }$ & $\frac { 100 } { 252 }$ & $a$ & $\frac { 25 } { 252 }$ & $\frac { 1 } { 252 }$ \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\roman*)]
\item State the value of $a$.
\item Hence write down the value of $\mathrm { E } ( X )$.
\item Determine the standard deviation of $X$.
\end{enumerate}\item Ali decides to play a game with Joanne using the same 10 balls. Joanne is asked to select 2 balls from the bag at random.
Ali agrees to pay Joanne 90 p if the two balls that she selects are the same colour, but nothing if they are different colours. Joanne pays 50 p to play the game.
The probability distribution of $Y$, the number of red balls that Joanne selects, is given in Table 2.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 2}
\begin{tabular}{ | c | c | c | c | }
\hline
$\boldsymbol { y }$ & 0 & 1 & 2 \\
\hline
$\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )$ & $\frac { 2 } { 9 }$ & $\frac { 5 } { 9 }$ & $\frac { 2 } { 9 }$ \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\roman*)]
\item Determine whether Joanne can expect to make a profit or a loss from playing the game once.
\item Hence calculate the expected size of this profit or loss after Joanne has played the game 100 times.\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2010 Q6 [10]}}