| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard Bayes with discrete events |
| Difficulty | Standard +0.3 This is a straightforward application of the law of total probability and Bayes' theorem with clearly defined events. Part (i) is routine probability tree calculation, part (ii) is direct Bayes formula application, and part (iii) involves basic binomial distribution work with some arithmetic. While it requires multiple steps, each component uses standard S4 techniques without requiring novel insight or complex problem-solving. |
| Spec | 2.03d Calculate conditional probability: from first principles5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(A) = P(K)\times 1 + P(K')\times\frac{1}{n}\) | M1 | |
| \(= p + (1-p)/n\) | A1 | allow \(p + \frac{q}{n}\) |
| \(= \frac{q+np}{n}\) AG | B1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(K \cap A) = p\) | B1 | |
| \(P(K | A) = \dfrac{p}{\dfrac{q+np}{n}}\) | M1 |
| \(= \dfrac{np}{q+np}\) | A1 | AEF |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(X\) answers are correct \(100-X\) are incorrect, so score \(= 2X-100 = 40\) giving \(X=70\) | B1 | 70 seen |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = 100 \times \frac{5}{8} = 62.5\) | B1 | |
| \(\text{Var}(X) = s^2 = 100 \times \frac{5}{8} \times \frac{3}{8} \ (= 23.4375) \ \left(= \frac{375}{16}\right)\) | M1A1 | Allow M1 from wrong \(p\) |
| \(P(X \geq 70) = 1 - \Phi\left[(69.5 - 62.5)/s\right]\) | M1A1 | Normal approximation. Allow M1 from \(40/70\) or wrong \(p\) |
| \(= 0.0741\) | A1 | Standardise M1 only if no or wrong cc, A1 for 0.0607 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(2X - 100) = 25\) | B1 | |
| \(\text{Var}(2X - 100) = 93.75\) | M1A1 | |
| \(P(2X - 100 \geq 40) = 1 - \Phi\left[(39 - 25)/\sqrt{93.75}\right]\) | M1A1 | Standardise, M1 only for no or wrong cc, A1 for 0.0671 |
| \(= 0.0741\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Score per question \(= S\) | ||
| \(E(S) = 1 \times \frac{5}{8} - 1 \times \frac{3}{8} = \frac{1}{4}\) | B1 | |
| \(\text{Var}(S) = 1^2 \times \frac{5}{8} + 1^2 \times \frac{3}{8} - \left(\frac{1}{4}\right)^2\) | M1A1 | |
| Total, \(T \sim N(25,\ 93.75)\) | ||
| \(P(T \geq 40) = 1 - \Phi\left[(39 - 25)/\sqrt{93.75}\right]\) | M1A1 | As for \(\beta\) |
| \(= 0.0741\) | B1 |
# Question 7:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(A) = P(K)\times 1 + P(K')\times\frac{1}{n}$ | M1 | |
| $= p + (1-p)/n$ | A1 | allow $p + \frac{q}{n}$ |
| $= \frac{q+np}{n}$ AG | B1 | |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(K \cap A) = p$ | B1 | |
| $P(K|A) = \dfrac{p}{\dfrac{q+np}{n}}$ | M1 | |
| $= \dfrac{np}{q+np}$ | A1 | AEF |
| **[3]** | | |
## Part (iii)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $X$ answers are correct $100-X$ are incorrect, so score $= 2X-100 = 40$ giving $X=70$ | B1 | 70 seen |
| **[1]** | | |
## Question 7(iii)(b):
$P(A) = \frac{5}{8}$
---
**Part (α):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 100 \times \frac{5}{8} = 62.5$ | B1 | |
| $\text{Var}(X) = s^2 = 100 \times \frac{5}{8} \times \frac{3}{8} \ (= 23.4375) \ \left(= \frac{375}{16}\right)$ | M1A1 | Allow M1 from wrong $p$ |
| $P(X \geq 70) = 1 - \Phi\left[(69.5 - 62.5)/s\right]$ | M1A1 | Normal approximation. Allow M1 from $40/70$ or wrong $p$ |
| $= 0.0741$ | A1 | Standardise M1 only if no or wrong cc, A1 for 0.0607 |
---
**Part (β):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(2X - 100) = 25$ | B1 | |
| $\text{Var}(2X - 100) = 93.75$ | M1A1 | |
| $P(2X - 100 \geq 40) = 1 - \Phi\left[(39 - 25)/\sqrt{93.75}\right]$ | M1A1 | Standardise, M1 only for no or wrong cc, A1 for 0.0671 |
| $= 0.0741$ | B1 | |
---
**Part (γ):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Score per question $= S$ | | |
| $E(S) = 1 \times \frac{5}{8} - 1 \times \frac{3}{8} = \frac{1}{4}$ | B1 | |
| $\text{Var}(S) = 1^2 \times \frac{5}{8} + 1^2 \times \frac{3}{8} - \left(\frac{1}{4}\right)^2$ | M1A1 | |
| Total, $T \sim N(25,\ 93.75)$ | | |
| $P(T \geq 40) = 1 - \Phi\left[(39 - 25)/\sqrt{93.75}\right]$ | M1A1 | As for $\beta$ |
| $= 0.0741$ | B1 | |
**Total: [6]**7 Each question on a multiple-choice examination paper has $n$ possible responses, only one of which is correct. Joni takes the paper and has probability $p$, where $0 < p < 1$, of knowing the correct response to any question, independently of any other. If she knows the correct response she will choose it, otherwise she will choose randomly from the $n$ possibilities. The events $K$ and $A$ are 'Joni knows the correct response' and 'Joni answers correctly' respectively.
\begin{enumerate}[label=(\roman*)]
\item Show that $\mathrm { P } ( A ) = \frac { q + n p } { n }$, where $q = 1 - p$.
\item Find $P ( K \mid A )$.
A paper with 100 questions has $n = 4$ and $p = 0.5$. Each correct response scores 1 and each incorrect response scores - 1 .
\item (a) Joni answers all the questions on the paper and scores 40 . How many questions did she answer correctly?\\
(b) By finding the distribution of the number of correct answers, or otherwise, find the probability that Joni scores at least 40 on the paper using her strategy.
\end{enumerate}
\hfill \mbox{\textit{OCR S4 2013 Q7 [13]}}