| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probability distribution from formula |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on discrete probability distributions. Part (i) requires summing a geometric series to find a constant (standard technique). Parts (ii)-(iv) involve basic probability calculations with given distributions. Parts (v)-(vi) apply the same concepts to a new context. All parts use routine A-level methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(x\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 16 } { 31 }\) | \(\frac { 8 } { 31 }\) | \(\frac { 4 } { 31 }\) | \(\frac { 2 } { 31 }\) | \(\frac { 1 } { 31 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(a(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}) = 1\) soi | M1 | AO 3.1a |
| \(a = \frac{16}{31}\) | A1 [2] | AO 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = 1, 3 \text{ or } 5) = \frac{21}{31}\) or \(0.677\) or \(0.68\) (2 sf) | B1 [1] | AO 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 2 \times \frac{21}{31} \times (1 - \frac{21}{31})\) | M1 | AO 2.1 |
| \(= \frac{420}{961}\) or \(0.437\) or \(0.44\) (2 sf) | A1 [2] | AO 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{2}{31} \times \frac{1}{31} + \frac{1}{31} \times \frac{2}{31}\) \((= \frac{4}{961})\) | M1 | AO 1.1a |
| \(\frac{P(\text{Sum}>8 \text{ & odd})}{P(\text{Sum odd})}\) | M1 | AO 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{105}\) or \(0.00952\) or \(0.0095\) (2 sf) | A1 [3] | AO 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_\infty = \frac{p}{1-0.5} = 1\) | M1 | AO 3.4 |
| \(P(X=1) = 0.5\) | A1 [2] | AO 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Eg \(Y\). (\(Y\) takes all values, but) \(X\) cannot be \(> 5\); Eg \(X\) because \(> 5\) is very unlikely | B1 [1] | AO 3.5b |
# Question 12:
## Part (i):
$a(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}) = 1$ soi | M1 | AO 3.1a | or $\frac{16}{31}(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}) = 1$ oe seen |
$a = \frac{16}{31}$ | A1 [2] | AO 1.1 | correctly obtained |
## Part (ii):
$P(X = 1, 3 \text{ or } 5) = \frac{21}{31}$ or $0.677$ or $0.68$ (2 sf) | B1 [1] | AO 1.1a |
## Part (iii):
$P(\text{sum odd}) = P(OE) + P(EO)$
$= 2 \times \frac{21}{31} \times (1 - \frac{21}{31})$ | M1 | AO 2.1 | or correct "long" method | Allow without "$2\times$" |
$= \frac{420}{961}$ or $0.437$ or $0.44$ (2 sf) | A1 [2] | AO 1.1 |
## Part (iv):
$P(\text{Sum} > 8 \text{ & odd}) = P(\text{Sum} = 9)$
$= P(4,5) + P(5,4)$
$= \frac{2}{31} \times \frac{1}{31} + \frac{1}{31} \times \frac{2}{31}$ $(= \frac{4}{961})$ | M1 | AO 1.1a | or $P(>8) \times P(O \mid > 8) = \frac{5}{961} \times \frac{4}{5}$ | Correct method |
$\frac{P(\text{Sum}>8 \text{ & odd})}{P(\text{Sum odd})}$ | M1 | AO 2.4 | Attempt ft their (iii) and their $P(\text{Sum} > 8$ & odd$)$ |
$= \frac{4}{961} \div \frac{420}{961}$
$= \frac{1}{105}$ or $0.00952$ or $0.0095$ (2 sf) | A1 [3] | AO 1.1 | cao **NB** $\frac{2}{961} \div \frac{210}{961} = \frac{1}{105}$ M0M1A0 |
## Part (v):
$S_\infty = \frac{p}{1-0.5} = 1$ | M1 | AO 3.4 |
$P(X=1) = 0.5$ | A1 [2] | AO 3.4 | Correct ans, no working M1A1 |
## Part (vi):
Eg $Y$. ($Y$ takes all values, but) $X$ cannot be $> 5$; Eg $X$ because $> 5$ is very unlikely | B1 [1] | AO 3.5b | oe, eg $Y$. It may take more than 5 attempts or "limited no." oe instead of 5 |
---12 The discrete random variable $X$ takes values 1, 2, 3, 4 and 5, and its probability distribution is defined as follows.
$$\mathrm { P } ( X = x ) = \begin{cases} a & x = 1 , \\ \frac { 1 } { 2 } \mathrm { P } ( X = x - 1 ) & x = 2,3,4,5 , \\ 0 & \text { otherwise } , \end{cases}$$
where $a$ is a constant.\\
(i) Show that $a = \frac { 16 } { 31 }$.
The discrete probability distribution for $X$ is given in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 16 } { 31 }$ & $\frac { 8 } { 31 }$ & $\frac { 4 } { 31 }$ & $\frac { 2 } { 31 }$ & $\frac { 1 } { 31 }$ \\
\hline
\end{tabular}
\end{center}
(ii) Find the probability that $X$ is odd.
Two independent values of $X$ are chosen, and their sum $S$ is found.\\
(iii) Find the probability that $S$ is odd.\\
(iv) Find the probability that $S$ is greater than 8 , given that $S$ is odd.
Sheila sometimes needs several attempts to start her car in the morning. She models the number of attempts she needs by the discrete random variable $Y$ defined as follows.
$$\mathrm { P } ( Y = y + 1 ) = \frac { 1 } { 2 } \mathrm { P } ( Y = y ) \quad \text { for all positive integers } y .$$
(v) Find $\mathrm { P } ( Y = 1 )$.\\
(vi) Give a reason why one of the variables, $X$ or $Y$, might be more appropriate as a model for the number of attempts that Sheila needs to start her car.
\hfill \mbox{\textit{OCR H240/02 2018 Q12 [11]}}