| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | 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 basic probability axioms (sum to 1), constructing a probability table, and simple conditional probability calculations. Part (a) uses ΣP(X=x)=1 to find k, parts (b-c) involve routine enumeration of outcomes with no conceptual challenges—easier than average A-level. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| \(k(1+2+3+4+5) = 1\) | M1 | Allow \(15k = 1\) |
| \(k = \frac{1}{15}\) | A1 | May be implied |
| \(P(X=3) = 3 \times \frac{1}{15}\) or \(\frac{3}{15}\) \((= 0.2\) AG) | A1 [3] | Must see \(3 \times \frac{1}{15}\) or \(\frac{3}{15}\) and answer \(0.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{15}, \frac{2}{15}, \frac{3}{15}, \frac{4}{15}, \frac{5}{15}\) oe; \(0.07, 0.13, 0.2, 0.27, 0.33\) | M1, A1 [2] | M1 for \(\geq 3\) probs correct, ft their \(k\); cao. Allow decimals (2 dp); SC: Table with all five probs \(= 0.2\): M1; Allow \(X = 0\) or \(X = 6\) or \(X = 6+\) if prob shown as 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3}{15} \times \frac{4}{15} + \frac{2}{15} \times \frac{5}{15}\) oe | M1 | Correct products added, ft their table |
| \(\times 2\) | M1 | \(2\times\)(Sum of two products of probs) |
| \(= \frac{44}{225}\) or \(0.196\) (3 sf) | A1 [3] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{one value is } 2 \text{ \& } T=7) = 2 \times \frac{2}{15} \times \frac{5}{15}\) | M1 | Allow without "\(2\times\)", ft their table |
| \(= \frac{4}{45}\) | A1f | ft their table (except if all probs \(= 0.2\)) |
| \(\frac{P(\text{one value is } 2 \text{ \& } T=7)}{P(T=7)} = \left(\frac{\frac{4}{45}}{\frac{44}{225}} \text{ or } \frac{0.0889}{0.196}\right)\) | M1 | Allow any probability / Their (c)(i) or their \(P(T=7)\) |
| \(= \frac{5}{11}\) or \(0.455\) (3 sf) | A1 [4] | cao NB not \(0.454\) |
# Question 12(a):
| $k(1+2+3+4+5) = 1$ | M1 | Allow $15k = 1$ |
| $k = \frac{1}{15}$ | A1 | May be implied |
| $P(X=3) = 3 \times \frac{1}{15}$ or $\frac{3}{15}$ $(= 0.2$ AG) | A1 [3] | Must see $3 \times \frac{1}{15}$ or $\frac{3}{15}$ and answer $0.2$ |
# Question 12(b):
| $\frac{1}{15}, \frac{2}{15}, \frac{3}{15}, \frac{4}{15}, \frac{5}{15}$ oe; $0.07, 0.13, 0.2, 0.27, 0.33$ | M1, A1 [2] | M1 for $\geq 3$ probs correct, ft their $k$; cao. Allow decimals (2 dp); SC: Table with all five probs $= 0.2$: M1; Allow $X = 0$ or $X = 6$ or $X = 6+$ if prob shown as 0 |
# Question 12(c)(i):
| $\frac{3}{15} \times \frac{4}{15} + \frac{2}{15} \times \frac{5}{15}$ oe | M1 | Correct products added, ft their table |
| $\times 2$ | M1 | $2\times$(Sum of two products of probs) |
| $= \frac{44}{225}$ or $0.196$ (3 sf) | A1 [3] | cao |
# Question 12(c)(ii):
| $P(\text{one value is } 2 \text{ \& } T=7) = 2 \times \frac{2}{15} \times \frac{5}{15}$ | M1 | Allow without "$2\times$", ft their table |
| $= \frac{4}{45}$ | A1f | ft their table (except if all probs $= 0.2$) |
| $\frac{P(\text{one value is } 2 \text{ \& } T=7)}{P(T=7)} = \left(\frac{\frac{4}{45}}{\frac{44}{225}} \text{ or } \frac{0.0889}{0.196}\right)$ | M1 | Allow any probability / Their (c)(i) or their $P(T=7)$ |
| $= \frac{5}{11}$ or $0.455$ (3 sf) | A1 [4] | cao NB not $0.454$ |
---12 A random variable $X$ has probability distribution defined as follows.
$$\mathrm { P } ( X = x ) = \begin{cases} k x & x = 1,2,3,4,5 , \\ 0 & \text { otherwise, } \end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 3 ) = 0.2$.
\item Show in a table the values of $X$ and their probabilities.
\item Two independent values of $X$ are chosen, and their total $T$ is found.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { P } ( T = 7 )$.
\item Given that $T = 7$, determine the probability that one of the values of $X$ is 2 .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2019 Q12 [12]}}