| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.8 This is a straightforward discrete probability distribution question requiring routine techniques: summing probabilities to find k, applying independence for joint probability, and using expected value with binomial distribution. All parts are standard textbook exercises with no novel insight required, making it easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k(1\times6+2\times5+3\times4+4\times3+5\times2+6\times1)=1\) oe | M1 (AO3.1a) | Allow one slip in arithmetic |
| \([k=]\frac{1}{56}\) isw | A1 (AO1.1) | B2 if unsupported |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((6\times k)^3\times(12\times k)^2\) oe seen | M1 (AO2.1) | FT their \(k\) |
| \(\frac{243}{4302592}\) or \(0.000056477584\) rounded to 2 or more sf | A1 (AO1.1) | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(40\times6k\) | M1 (AO3.1b) | FT their \(k\) |
| \(4.286\) or \(4.29\) or \(4.3\) | A1 (AO3.2b) | Mark the final answer |
| [2] |
## Question 11(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k(1\times6+2\times5+3\times4+4\times3+5\times2+6\times1)=1$ oe | M1 (AO3.1a) | Allow one slip in arithmetic |
| $[k=]\frac{1}{56}$ isw | A1 (AO1.1) | B2 if unsupported |
| **[2]** | | |
## Question 11(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(6\times k)^3\times(12\times k)^2$ oe seen | M1 (AO2.1) | FT their $k$ |
| $\frac{243}{4302592}$ or $0.000056477584$ rounded to 2 or more sf | A1 (AO1.1) | |
| **[2]** | | |
## Question 11(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $40\times6k$ | M1 (AO3.1b) | FT their $k$ |
| $4.286$ or $4.29$ or $4.3$ | A1 (AO3.2b) | Mark the final answer |
| **[2]** | | |
---11 The discrete random variable $X$ takes the values $0,1,2,3,4$ and 5 with probabilities given by the formula
$$\mathrm { P } ( X = x ) = k ( x + 1 ) ( 6 - x ) .$$
(i) Find the value of $k$.
In one half-term Ben attends school on 40 days. The probability distribution above is used to model $X$, the number of lessons per day in which Ben receives a gold star for excellent work.\\
(ii) Find the probability that Ben receives no gold stars on each of the first 3 days of the half-term and two gold stars on each of the next 2 days.\\
(iii) Find the expected number of days in the half-term on which Ben receives no gold stars.
\hfill \mbox{\textit{OCR MEI Paper 2 2018 Q11 [6]}}