| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sequential trials until success |
| Difficulty | Standard +0.8 This S2 question requires careful enumeration of sequential trial outcomes with two stopping conditions, probability calculations with different biases, and a multi-stage problem involving matching distributions and binomial probability. The conceptual challenge of tracking 'success or 5 failures' across different probability models, plus the compound probability in part (c), elevates this above routine S2 exercises but remains within standard A-level scope. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(\boldsymbol { n }\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathbf { P } ( \boldsymbol { N } = \boldsymbol { n } )\) | \(\frac { 1 } { 8 }\) | \(\frac { 1 } { 16 }\) |
| \(\boldsymbol { m }\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathbf { P } ( \boldsymbol { M } = \boldsymbol { m } )\) | \(\frac { 1 } { 4 }\) | \(\frac { 3 } { 16 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(n\) | Outcome | \(P(N = n)\) |
| 1 | H | \(\mathbf{0.5}\) (\(\frac{1}{2}\)) |
| 2 | TH | \(\mathbf{0.25}\) (\(\frac{1}{4}\)) |
| 3 | TTH | \(\mathbf{0.125}\) (\(\frac{1}{8}\)) |
| 4 | TTTH | \(\mathbf{0.0625}\) (\(\frac{1}{16}\)) |
| 5 | TTTTTA | \(\mathbf{0.0625}\) (\(\frac{1}{16}\)) |
| \(E(N) = \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{4}\right) + \left(3 \times \frac{1}{8}\right) + \left(4 \times \frac{1}{16}\right) + \left(5 \times \frac{1}{16}\right) = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{16} = \frac{31}{16} = 1.9375\) | M1 A1 | \(\sum_{n=1}^{5} n \times P(N = n)\) (all 5 terms attempted /seen/ implied) (awfw 1.93 to 1.94) |
| Answer | Marks | Guidance |
|---|---|---|
| \(m\) | Outcome | \(P(M = m)\) |
| 1 | H | \(\frac{1}{4}\) |
| 2 | TH | \(\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\) |
| 3 | TTH | \(\left(\frac{1}{4}\right)^2 \times \frac{1}{4} = \frac{9}{64}\) |
| 4 | TTTH | \(\left(\frac{1}{4}\right)^3 \times \frac{1}{4} = \frac{27}{256}\) |
| 5 | TTTTTA | \(\left(\frac{1}{4}\right)^4 \times 1 = \frac{81}{256}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(J,R):\) | M1 | e.g 0.125 attempt at any \(P(n,m)\) |
| \(P(1,1) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\) (oe) | A1 | any 1 correct to 3sf |
| \(P(2,2) = \frac{1}{4} \times \frac{3}{16} = \frac{3}{64}\) (oe) | A1 | all 5 correct to 3sf |
| \(P(3,3) = \frac{1}{8} \times \frac{9}{64} = \frac{9}{512}\) (oe) | ||
| \(P(4,4) = \frac{1}{16} \times \frac{27}{256} = \frac{27}{4096}\) (oe) | ||
| \(P(5,5) = \frac{1}{16} \times \frac{81}{256} = \frac{81}{4096}\) (oe) | ||
| \(p = \sum P(n,n) \Rightarrow p = \frac{221}{1024} (0.2158)\) | ml A1 | \(\sum_{n=1}^{5} P(n,n)\) with all 5 values attempted (awfw 0.215 to 0.217) (either term with their \(p\) used) (\(0 < p < 1\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 3 \times \left(\frac{221}{1024}\right)^2 \times \left(\frac{803}{1024}\right)\) | M1 | (second term with their \(p\) used) (\(0 < p < 1\)) |
| \(+ \left(\frac{221}{1024}\right)^3\) | M1 | |
| \(P(X \geq 2) = P(X = 2) + P(X = 3) = 0.120\) (3dp) | Mdep1 A1 | dep (M1M1) (allow 0.119; 0.12; 0.121) |
## (a)
| $n$ | Outcome | $P(N = n)$ |
|---|---|---|
| 1 | H | $\mathbf{0.5}$ ($\frac{1}{2}$) |
| 2 | TH | $\mathbf{0.25}$ ($\frac{1}{4}$) |
| 3 | TTH | $\mathbf{0.125}$ ($\frac{1}{8}$) |
| 4 | TTTH | $\mathbf{0.0625}$ ($\frac{1}{16}$) |
| 5 | TTTTTA | $\mathbf{0.0625}$ ($\frac{1}{16}$) | B2,1 | B1 for one correct entry for $n = 1, 2, 4$ B2 for all 3 correct |
| $E(N) = \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{4}\right) + \left(3 \times \frac{1}{8}\right) + \left(4 \times \frac{1}{16}\right) + \left(5 \times \frac{1}{16}\right) = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{16} = \frac{31}{16} = 1.9375$ | M1 A1 | $\sum_{n=1}^{5} n \times P(N = n)$ (all 5 terms attempted /seen/ implied) (awfw 1.93 to 1.94) |
## (b)
| $m$ | Outcome | $P(M = m)$ |
|---|---|---|
| 1 | H | $\frac{1}{4}$ |
| 2 | TH | $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$ |
| 3 | TTH | $\left(\frac{1}{4}\right)^2 \times \frac{1}{4} = \frac{9}{64}$ |
| 4 | TTTH | $\left(\frac{1}{4}\right)^3 \times \frac{1}{4} = \frac{27}{256}$ |
| 5 | TTTTTA | $\left(\frac{1}{4}\right)^4 \times 1 = \frac{81}{256}$ | B3,2,1 | (B1 any one correct) (B2 any 2 correct) (B3 all 3 correct) |
## (c)(i)
| $P(J,R):$ | M1 | e.g 0.125 attempt at any $P(n,m)$ |
| $P(1,1) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ (oe) | A1 | any 1 correct to 3sf |
| $P(2,2) = \frac{1}{4} \times \frac{3}{16} = \frac{3}{64}$ (oe) | A1 | all 5 correct to 3sf |
| $P(3,3) = \frac{1}{8} \times \frac{9}{64} = \frac{9}{512}$ (oe) | | |
| $P(4,4) = \frac{1}{16} \times \frac{27}{256} = \frac{27}{4096}$ (oe) | | |
| $P(5,5) = \frac{1}{16} \times \frac{81}{256} = \frac{81}{4096}$ (oe) | | |
| $p = \sum P(n,n) \Rightarrow p = \frac{221}{1024} (0.2158)$ | ml A1 | $\sum_{n=1}^{5} P(n,n)$ with all 5 values attempted (awfw 0.215 to 0.217) (either term with their $p$ used) ($0 < p < 1$) |
## (c)(ii)
| $= 3 \times \left(\frac{221}{1024}\right)^2 \times \left(\frac{803}{1024}\right)$ | M1 | (second term with their $p$ used) ($0 < p < 1$) |
| $+ \left(\frac{221}{1024}\right)^3$ | M1 | |
| $P(X \geq 2) = P(X = 2) + P(X = 3) = 0.120$ (3dp) | Mdep1 A1 | dep (M1M1) (allow 0.119; 0.12; 0.121) |
**Total: 16 marks**
---5
\begin{enumerate}[label=(\alph*)]
\item Joshua plays a game in which he repeatedly tosses an unbiased coin. His game concludes when he obtains either a head or 5 tails in succession.
The random variable $N$ denotes the number of tosses of his coin required to conclude a game.
By completing Table 3 below, calculate $\mathrm { E } ( N )$.
\item Joshua's sister, Ruth, plays a separate game in which she repeatedly tosses a coin that is biased in such a way that the probability of a head in a single toss of her coin is $\frac { 1 } { 4 }$. Her game also concludes when she obtains either a head or 5 tails in succession.
The random variable $M$ denotes the number of tosses of her coin required to conclude her game.
Complete Table 4 below.
\item \begin{enumerate}[label=(\roman*)]
\item Joshua and Ruth play their games simultaneously. Calculate the probability that Joshua and Ruth will conclude their games in an equal number of tosses of their coins.
\item Joshua and Ruth play their games simultaneously on 3 occasions. Calculate the probability that, on at least 2 of these occasions, their games will be concluded in an equal number of tosses of their coins. Give your answer to three decimal places.\\
(4 marks)
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 3}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$\boldsymbol { n }$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathbf { P } ( \boldsymbol { N } = \boldsymbol { n } )$ & & & $\frac { 1 } { 8 }$ & & $\frac { 1 } { 16 }$ \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 4}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$\boldsymbol { m }$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathbf { P } ( \boldsymbol { M } = \boldsymbol { m } )$ & $\frac { 1 } { 4 }$ & $\frac { 3 } { 16 }$ & & & \\
\hline
\end{tabular}
\end{center}
\end{table}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q5 [16]}}