| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Comparison or ordering of two independent values |
| Difficulty | Challenging +1.2 This question requires finding p from probability axioms, systematically enumerating cases where xy > x+y, and understanding sampling with/without replacement. Part (b) involves symmetry reasoning with equally likely outcomes. While multi-step, the techniques are standard A-level probability methods without requiring novel insight or complex algebraic manipulation. |
| Spec | 2.03a Mutually exclusive and independent events2.04a Discrete probability distributions |
| \(x\) | 0 | 1 | 2 | 3 |
| \(P ( X = x )\) | \(\frac { 1 } { 12 }\) | \(\frac { 1 } { 4 }\) | \(p\) | \(3 p\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{12} + \frac{1}{4} + p + 3p = 1\) | M1 | |
| \(p = \frac{1}{6}\) | A1 | |
| \((2,3)\ (3,2)\ (3,3)\) | M1 | soi. Allow M1 for any two correct pairs and no incorrect pairs, or for all three correct pairs and at most one incorrect pair |
| \(\frac{1}{6} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{2}\) | M1 | All correct and added, ft their \(p\) |
| \(\frac{5}{12}\) | A1, [5] |
| Answer | Marks |
|---|---|
| \(\frac{1}{2}(1 - 0.02)\) | M1 |
| \(= 0.49\) | A1, [2] |
# Question 13:
## Part (a):
$\frac{1}{12} + \frac{1}{4} + p + 3p = 1$ | M1 |
$p = \frac{1}{6}$ | A1 |
$(2,3)\ (3,2)\ (3,3)$ | M1 | soi. Allow M1 for any two correct pairs and no incorrect pairs, or for all three correct pairs and at most one incorrect pair
$\frac{1}{6} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{2}$ | M1 | All correct and added, ft their $p$
$\frac{5}{12}$ | A1, [5] |
## Part (b):
$\frac{1}{2}(1 - 0.02)$ | M1 |
$= 0.49$ | A1, [2] |13
\begin{enumerate}[label=(\alph*)]
\item The probability distribution of a random variable $X$ is shown in the table, where $p$ is a constant.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$P ( X = x )$ & $\frac { 1 } { 12 }$ & $\frac { 1 } { 4 }$ & $p$ & $3 p$ \\
\hline
\end{tabular}
\end{center}
Two values of $X$ are chosen at random. Determine the probability that their product is greater than their sum.
\item A random variable $Y$ takes $n$ values, each of which is equally likely. Two values, $Y _ { 1 }$ and $Y _ { 2 }$, of $Y$ are chosen at random.
It is given that $\mathrm { P } \left( Y _ { 1 } = Y _ { 2 } \right) = 0.02$.\\
Find $\mathrm { P } \left( Y _ { 1 } > Y _ { 2 } \right)$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q13 [7]}}