| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Moderate -0.8 This is a straightforward binomial distribution question requiring direct application of standard formulas. Part (i) involves routine calculations of P(X=6), P(X<6), and finding E(X) and Var(X) using n=9, p=2/3. Part (ii) requires recognizing that X₁+X₂+X₃ follows B(27, 2/3), which is a small conceptual step but still standard. All techniques are textbook exercises with no novel problem-solving required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Binomial seen or implied | M1 | by use of table or \(^9C_6\) or \((\frac{2}{3})^p(\frac{1}{3})^q\) \((p+q=9)\) — Eg 0.6228 seen |
| \(0.6228 - 0.3497\) | M1 | \(^9C_6(\frac{1}{3})^3(\frac{2}{3})^6\) |
| \(= 0.273\) (3 sf) | A1 | \(\frac{1792}{6561}\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3497\) or \(0.350\) (3 sf) | B1 | NB 0.3498 (from 0.6228 - 0.273) rounds to 0.350 so B1 |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 6 | B1ft | |
| 2 | B1ft | NB 2, 6 B0B0 unless labelled correctly |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 27 seen | B1 | not necessarily in a statement |
| \(B(27, \frac{2}{3})\) seen or implied | M1 | |
| \(^{27}C_{18}(\frac{1}{3})^9(\frac{2}{3})^{18}\) | M1 | or attempt eg \(P(X_1=1)\times P(X_2=8)\times P(X_3=9)\), \(P(X_1=2)\times P(X_2=7)\times P(X_3=9)\), \(P(X_1=3)\times P(X_2=6)\times P(X_3=9)\) etc \(\geq 3\) sets with \(X_1+X_2+X_3=18\) (not nec'y added) M1 — NB \(P(X_1=6)\times P(X_2=6)\times P(X_3=6) = 0.273^3 = 0.0203\) M0M0A0; \(\frac{55}{729}\ (=0.0754)\) M0M0A0 |
| \(= 0.161\) (3 sf) | A1 | |
| [4] |
# Question 4:
**Note:** If $\frac{2}{3}$ is interpreted consistently as 0.6 or 0.66 or 0.67 or 0.7, max marks: (i)(a) M1M1A0, (i)(b) B0, (i)(c) B1ft B1ft, (ii) B1M1M1A0
## Part 4(i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Binomial seen or implied | M1 | by use of table or $^9C_6$ or $(\frac{2}{3})^p(\frac{1}{3})^q$ $(p+q=9)$ — Eg 0.6228 seen |
| $0.6228 - 0.3497$ | M1 | $^9C_6(\frac{1}{3})^3(\frac{2}{3})^6$ |
| $= 0.273$ (3 sf) | A1 | $\frac{1792}{6561}$ |
| **[3]** | | |
## Part 4(i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3497$ or $0.350$ (3 sf) | B1 | NB 0.3498 (from 0.6228 - 0.273) rounds to 0.350 so B1 |
| **[1]** | | |
## Part 4(i)(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 6 | B1ft | |
| 2 | B1ft | NB 2, 6 B0B0 unless labelled correctly |
| **[2]** | | |
## Part 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 27 seen | B1 | not necessarily in a statement |
| $B(27, \frac{2}{3})$ seen or implied | M1 | |
| $^{27}C_{18}(\frac{1}{3})^9(\frac{2}{3})^{18}$ | M1 | or attempt eg $P(X_1=1)\times P(X_2=8)\times P(X_3=9)$, $P(X_1=2)\times P(X_2=7)\times P(X_3=9)$, $P(X_1=3)\times P(X_2=6)\times P(X_3=9)$ etc $\geq 3$ sets with $X_1+X_2+X_3=18$ (not nec'y added) M1 — NB $P(X_1=6)\times P(X_2=6)\times P(X_3=6) = 0.273^3 = 0.0203$ M0M0A0; $\frac{55}{729}\ (=0.0754)$ M0M0A0 |
| $= 0.161$ (3 sf) | A1 | |
| **[4]** | | |
---4 Each time Ben attempts to complete a crossword in his daily newspaper, the probability that he succeeds is $\frac { 2 } { 3 }$. The random variable $X$ denotes the number of times that Ben succeeds in 9 attempts.\\
(i) Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( X = 6 )$,
\item $\mathrm { P } ( X < 6 )$,
\item $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
Ben notes three values, $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$, of $X$.\\
(ii) State the total number of attempts to complete a crossword that are needed to obtain three values of $X$. Hence find $\mathrm { P } \left( X _ { 1 } + X _ { 2 } + X _ { 3 } = 18 \right)$.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2014 Q4 [10]}}