| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Sum or combination of independent binomial values |
| Difficulty | Standard +0.8 Part (i) is routine binomial calculation. Part (ii)(a) is standard. Part (ii)(b) requires recognizing that the sum of three independent B(4,0.7) variables follows B(12,0.7), then calculating P(total=10) - this conceptual leap about sums of binomial variables is non-standard for S1 and requires insight beyond typical textbook exercises. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - 0.1754\) alone \(= 0.825\) (3 sfs) | M1, A1 | Allow \(1 - 0.2855\) or \(0.7145\) or \(0.715\) alone |
| Answer | Marks | Guidance |
|---|---|---|
| \(^4C_2 \times 0.7^2 \times 0.3^2 = \frac{1323}{5000}\) or \(0.265\) (3 sf) | M1, A1 | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| 4,4,2 & 4,3,3 only, seen or implied | B1 | Both needed |
| \(P(Y=4) = 0.7^4\) (or \(\frac{2401}{10000}\) or 0.2401) | M1 | |
| \(P(Y=3) = 4 \times 0.3 \times 0.7^3\) (or \(\frac{1029}{2500}\) or 0.4116) | M1 | |
| \(P(4,3,3) = 3 \times "0.2401" \times "0.4116"^2\) (or 0.122) | M1 | ie \(3 \times\) their \(P(4) \times (\text{their } P(3))^2\); if "3×" omitted twice or "3!×" used twice allow M1M0 |
| \(P(4,4,2) = 3 \times 0.2401"^2 \times "0.265"\) (or 0.0458) | M1 | ie \(3 \times (\text{their } P(4))^2 \times \text{their } P(2)\) ft (ii)(a); For M mks ignore extra combs eg P(4,4,3); eg ans 0.0560, 0.0559, 0.336, probably B1M1M1M1M1M0A0 but must see method |
| \(P(\text{Tot} = 10) = 0.168\) (3 sfs) | A1 | If B(30, 0.6) clearly being used: Any 5 combs adding to 10 seen B1; \(P(8) = {^{30}C_8} \times 0.4^{22} \times 0.6^8\) or 0.0002; \(P(9) = {^{30}C_9} \times 0.4^{21} \times 0.6^9\) or 0.0007; \(P(10) = {^{30}C_{10}} \times 0.4^{20} \times 0.6^{10}\) or 0.0020; all three correct M2 or two correct M1; No more marks |
# Question 8(i):
$1 - 0.1754$ alone $= 0.825$ (3 sfs) | M1, A1 | Allow $1 - 0.2855$ or $0.7145$ or $0.715$ alone
**Total: [2]**
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# Question 8(ii)(a):
$^4C_2 \times 0.7^2 \times 0.3^2 = \frac{1323}{5000}$ or $0.265$ (3 sf) | M1, A1 | All correct
**Total: [2]**
---
# Question 8(ii)(b):
4,4,2 & 4,3,3 only, seen or implied | B1 | Both needed
$P(Y=4) = 0.7^4$ (or $\frac{2401}{10000}$ or 0.2401) | M1 |
$P(Y=3) = 4 \times 0.3 \times 0.7^3$ (or $\frac{1029}{2500}$ or 0.4116) | M1 |
$P(4,3,3) = 3 \times "0.2401" \times "0.4116"^2$ (or 0.122) | M1 | ie $3 \times$ their $P(4) \times (\text{their } P(3))^2$; if "3×" omitted twice or "3!×" used twice allow M1M0
$P(4,4,2) = 3 \times 0.2401"^2 \times "0.265"$ (or 0.0458) | M1 | ie $3 \times (\text{their } P(4))^2 \times \text{their } P(2)$ ft (ii)(a); For M mks ignore extra combs eg P(4,4,3); eg ans 0.0560, 0.0559, 0.336, probably B1M1M1M1M1M0A0 but must see method
$P(\text{Tot} = 10) = 0.168$ (3 sfs) | A1 | If B(30, 0.6) clearly being used: Any 5 combs adding to 10 seen B1; $P(8) = {^{30}C_8} \times 0.4^{22} \times 0.6^8$ or 0.0002; $P(9) = {^{30}C_9} \times 0.4^{21} \times 0.6^9$ or 0.0007; $P(10) = {^{30}C_{10}} \times 0.4^{20} \times 0.6^{10}$ or 0.0020; all three correct M2 or two correct M1; No more marks
**Total: [6]**
---8 (i) The random variable $X$ has the distribution $\mathrm { B } ( 30,0.6 )$. Find $\mathrm { P } ( X \geqslant 16 )$.\\
(ii) The random variable $Y$ has the distribution $\mathrm { B } ( 4,0.7 )$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( Y = 2 )$.
\item Three values of $Y$ are chosen at random. Find the probability that their total is 10 .
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2012 Q8 [10]}}