| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of binomial probability with p=0.5, requiring only direct calculation using the formula or tables. Part (i) is a simple tail probability, part (ii)(a) uses symmetry for an even number of trials, and part (ii)(b) exploits the symmetry of the binomial distribution with p=0.5. No problem-solving insight is needed beyond recognizing the standard binomial setup and using basic symmetry properties. |
| Spec | 5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Binomial stated | M1 | or implied by \(C \times 0.5^r\) or use of table; or \(0.5^7{\times}0.5 + 0.5^8\) or \(0.5^8 + 0.5^8\) |
| \(1 - 0.9648\) | M1 | or \({}^8C_7 \times 0.5^7{\times}0.5 + 0.5^8\) fully correct method; \(1-(0.5^8+8{\times}0.5^8+{}^8C_2 0.5^8\ldots)\) all correct |
| \(= 0.0352\) (3 sfs) or \(\frac{9}{256}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \({}^{22}C_{11} \times 0.5^{11} \times 0.5^{11}\) | M1 | Fully correct method. Not ISW |
| \(= 0.168\) (3 sfs) | A1 | e.g. \(0.168^2\) or \(2 \times 0.168\) or \(1-0.168\): M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - \text{"0.168"}\) | M1 | or \(0.5^{22}({}^{22}C_{12} + {}^{22}C_{13} + {}^{22}C_{14} + \ldots + 22 + 1)\); All 11 correct terms seen, or correct ans: M2 |
| \(\frac{1}{2}(1 - \text{"0.168"})\) | M1 | or \(P(X = 12, 13, \ldots 21, 22)\) stated or implied with \(\geq 2\) terms shown or one extra term |
| \(= 0.416\) (3 sfs) | A1 |
## Question 8:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Binomial stated | M1 | or implied by $C \times 0.5^r$ or use of table; or $0.5^7{\times}0.5 + 0.5^8$ or $0.5^8 + 0.5^8$ |
| $1 - 0.9648$ | M1 | or ${}^8C_7 \times 0.5^7{\times}0.5 + 0.5^8$ fully correct method; $1-(0.5^8+8{\times}0.5^8+{}^8C_2 0.5^8\ldots)$ all correct |
| $= 0.0352$ (3 sfs) or $\frac{9}{256}$ | A1 | |
### Part (ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| ${}^{22}C_{11} \times 0.5^{11} \times 0.5^{11}$ | M1 | Fully correct method. Not ISW |
| $= 0.168$ (3 sfs) | A1 | e.g. $0.168^2$ or $2 \times 0.168$ or $1-0.168$: M0A0 |
### Part (ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - \text{"0.168"}$ | M1 | or $0.5^{22}({}^{22}C_{12} + {}^{22}C_{13} + {}^{22}C_{14} + \ldots + 22 + 1)$; All 11 correct terms seen, or correct ans: M2 |
| $\frac{1}{2}(1 - \text{"0.168"})$ | M1 | or $P(X = 12, 13, \ldots 21, 22)$ stated or implied with $\geq 2$ terms shown or one extra term |
| $= 0.416$ (3 sfs) | A1 | |
---8 On average, half the plants of a particular variety produce red flowers and the rest produce blue flowers.\\
(i) Ann chooses 8 plants of this variety at random. Find the probability that more than 6 plants produce red flowers.\\
(ii) Karim chooses 22 plants of this variety at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the number of these plants that produce blue flowers is equal to the number that produce red flowers.
\item Hence find the probability that the number of these plants that produce blue flowers is greater than the number that produce red flowers.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2012 Q8 [8]}}