| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probability distribution from formula |
| Difficulty | Standard +0.8 This question progresses from routine probability distribution setup (parts a-b) to challenging multi-step probability calculations. Part (c) requires systematic enumeration of cases where X₁ > X₂ + X₃, and part (d) demands careful consideration of all possible sequences summing to exactly 7 on the 5th observation without hitting 7 earlier—a non-trivial combinatorial problem requiring organized case work and independence calculations. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) = 1\) | M1 | Correct equation involving multiple of \(k\) |
| \(k \times \frac{25}{12} = 1\) or eg \(\frac{25}{3}k = 4\) or \(25k = 12\) | ||
| hence \(k = \frac{12}{25}\) AG | A1 | Must see previous line and answer |
| or verification: \(\frac{12}{25} + \frac{6}{25} + \frac{4}{25} + \frac{3}{25} = 1\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x\): 1, 2, 3, 4 with \(P(X=x)\): \(\frac{12}{25},\ \frac{6}{25},\ \frac{4}{25},\ \frac{3}{25}\) | B1 | Or equivalent exact values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((3,1,1)\ (4,1,1)\ (4,2,1)\ (4,1,2)\) | M1 | At least three of these seen or implied. No extras or repeats. |
| \(\frac{4}{25}\times\left(\frac{12}{25}\right)^2 + \frac{3}{25}\times\left(\frac{12}{25}\right)^2 + \frac{3}{25}\times\frac{6}{25}\times\frac{12}{25} + \frac{3}{25}\times\frac{12}{25}\times\frac{6}{25}\) | M1 | At least two correct terms, no incorrect coefficients; ft their table. Allow in terms of \(k\). |
| \(= \frac{288}{3125}\) or \(0.09216\) | A1 | Allow \(0.0922\) (3 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((1,1,1,1,3)\) and \((1,1,1,2,2)\) | B1, B1 | B1B1 for both sets in any order, without extras. Both soi. B1 for both sets in any order, with extras. |
| \(\left(\frac{12}{25}\right)^4\times\frac{4}{25}\times 5 + \left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\times\ ^5C_2\) | M1 | \(\left(\frac{12}{25}\right)^4\times\frac{4}{25}\) or \(\left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\) seen. Ignore coefficients. ft their table. |
| A1 | For either \(\left(\frac{12}{25}\right)^4\times\frac{4}{25}\times 5\) or \(\left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\times\ ^5C_2\) oe ft their table | |
| Alternative method: \(\left(\frac{12}{25}\right)^4\times\frac{4}{25}\times(4+1)+\left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\times(^4C_2+4)\) | M1 | oe |
| A1 | For either \(\left(\frac{12}{25}\right)^4\times\frac{4}{25}\times(4+1)\) or \(\left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\times(^4C_2+4)\) oe | |
| \(= \frac{41472}{390625}\) or \(0.10616832\) | A1 | Allow \(0.106\) (3 sf) |
# Question 14:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $k\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) = 1$ | M1 | Correct equation involving multiple of $k$ |
| $k \times \frac{25}{12} = 1$ or eg $\frac{25}{3}k = 4$ or $25k = 12$ | | |
| hence $k = \frac{12}{25}$ **AG** | A1 | Must see previous line and answer |
| or verification: $\frac{12}{25} + \frac{6}{25} + \frac{4}{25} + \frac{3}{25} = 1$ | M1, A1 | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x$: 1, 2, 3, 4 with $P(X=x)$: $\frac{12}{25},\ \frac{6}{25},\ \frac{4}{25},\ \frac{3}{25}$ | B1 | Or equivalent exact values |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(3,1,1)\ (4,1,1)\ (4,2,1)\ (4,1,2)$ | M1 | At least three of these seen or implied. No extras or repeats. |
| $\frac{4}{25}\times\left(\frac{12}{25}\right)^2 + \frac{3}{25}\times\left(\frac{12}{25}\right)^2 + \frac{3}{25}\times\frac{6}{25}\times\frac{12}{25} + \frac{3}{25}\times\frac{12}{25}\times\frac{6}{25}$ | M1 | At least two correct terms, no incorrect coefficients; ft their table. Allow in terms of $k$. |
| $= \frac{288}{3125}$ or $0.09216$ | A1 | Allow $0.0922$ (3 sf) |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(1,1,1,1,3)$ and $(1,1,1,2,2)$ | B1, B1 | B1B1 for both sets in any order, without extras. Both soi. B1 for both sets in any order, with extras. |
| $\left(\frac{12}{25}\right)^4\times\frac{4}{25}\times 5 + \left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\times\ ^5C_2$ | M1 | $\left(\frac{12}{25}\right)^4\times\frac{4}{25}$ or $\left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2$ seen. Ignore coefficients. ft their table. |
| | A1 | For either $\left(\frac{12}{25}\right)^4\times\frac{4}{25}\times 5$ or $\left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\times\ ^5C_2$ oe ft their table |
| **Alternative method:** $\left(\frac{12}{25}\right)^4\times\frac{4}{25}\times(4+1)+\left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\times(^4C_2+4)$ | M1 | oe |
| | A1 | For either $\left(\frac{12}{25}\right)^4\times\frac{4}{25}\times(4+1)$ or $\left(\frac{12}{25}\right)^3\times\left(\frac{6}{25}\right)^2\times(^4C_2+4)$ oe |
| $= \frac{41472}{390625}$ or $0.10616832$ | A1 | Allow $0.106$ (3 sf) |14 The probability distribution of a random variable $X$ is modelled as follows.\\
$\mathrm { P } ( X = x ) = \begin{cases} \frac { k } { x } & x = 1,2,3,4 , \\ 0 & \text { otherwise, } \end{cases}$\\
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 12 } { 25 }$.
\item Show in a table the values of $X$ and their probabilities.
\item The values of three independent observations of $X$ are denoted by $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$.
Find $\mathrm { P } \left( X _ { 1 } > X _ { 2 } + X _ { 3 } \right)$.
In a game, a player notes the values of successive independent observations of $X$ and keeps a running total. The aim of the game is to reach a total of exactly 7 .
\item Determine the probability that a total of exactly 7 is first reached on the 5th observation.
\section*{OCR}
Oxford Cambridge and RSA
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2021 Q14 [11]}}