| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.8 This is a very routine probability distribution question requiring only the basic principle that probabilities sum to 1. Part (a) involves simple algebra (0.2 + 0.1 + k + 2k + 4k = 1, solve for k), and part (b) is straightforward complement/subtraction. No problem-solving or conceptual depth required—pure mechanical application of fundamental definitions. |
| Spec | 5.02a Discrete probability distributions: general |
| \(x\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( X = x )\) | 0.2 | 0.1 | \(k\) | \(2 k\) | \(4 k\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.2 + 0.1 + k + 2k + 4k [= 1]\) soi | M1 (AO 1.1a) | Sum of probabilities set equal to 1 |
| \(k = 0.1\) | A1 (AO 1.1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - 2 \times \text{their } k\) | M1 (AO 1.1) | \(1 - \text{their } P(X=4)\) oe |
| \(0.8\) or \(\frac{4}{5}\) cao | A1 (AO 1.1) |
## Question 1:
### Part (a)
$0.2 + 0.1 + k + 2k + 4k [= 1]$ soi | M1 (AO 1.1a) | Sum of probabilities set equal to 1
$k = 0.1$ | A1 (AO 1.1) |
**[2 marks]**
### Part (b)
$1 - 2 \times \text{their } k$ | M1 (AO 1.1) | $1 - \text{their } P(X=4)$ oe
$0.8$ or $\frac{4}{5}$ cao | A1 (AO 1.1) |
**[2 marks]**
---1 Fig. 1 shows the probability distribution of the discrete random variable $X$.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = x )$ & 0.2 & 0.1 & $k$ & $2 k$ & $4 k$ \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find $\mathrm { P } ( X \neq 4 )$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2019 Q1 [4]}}