| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Moderate -0.3 This is a straightforward application of probability axioms (sum to 1) and expectation formula to set up two simultaneous equations, followed by routine algebraic solving. Part (ii) requires the standard variance formula. All steps are mechanical with no conceptual challenges beyond basic S1 definitions. |
| Spec | 2.04a Discrete probability distributions |
| \(r\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = r )\) | 0.5 | 0.35 | \(p\) | \(q\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((A)\): \(0.5 + 0.35 + p + q = 1\), so \(p + q = 0.15\) | B1 | \(p + q\) in a correct equation before reaching \(p+q=0.15\) |
| \((B)\): \(0\times0.5 + 1\times0.35 + 2p + 3q = 0.67\), so \(2p + 3q = 0.32\) | B1 | \(2p + 3q\) in a correct equation before reaching \(2p+3q=0.32\) |
| \((C)\): from above \(2p + 2q = 0.30\), so \(q = 0.02,\ p = 0.13\) | B1 | for any 1 correct answer |
| B2 | for both correct answers — Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X^2) = 0\times0.5 + 1\times0.35 + 4\times0.13 + 9\times0.02 = 1.05\) | M1 | \(\Sigma x^2 p\) (at least 2 non-zero terms correct) |
| \(\text{Var}(X) = \text{'their } 1.05\text{'} - 0.67^2 = 0.6011\) (awrt 0.6) | M1dep | for \((-0.67^2)\), provided \(\text{Var}(X) > 0\) |
| A1 | cao (No \(n\) or \(n-1\) divisors) — Total: 3 |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(A)$: $0.5 + 0.35 + p + q = 1$, so $p + q = 0.15$ | B1 | $p + q$ in a correct equation before reaching $p+q=0.15$ |
| $(B)$: $0\times0.5 + 1\times0.35 + 2p + 3q = 0.67$, so $2p + 3q = 0.32$ | B1 | $2p + 3q$ in a correct equation before reaching $2p+3q=0.32$ |
| $(C)$: from above $2p + 2q = 0.30$, so $q = 0.02,\ p = 0.13$ | B1 | for any 1 correct answer |
| | B2 | for both correct answers — **Total: 2** |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X^2) = 0\times0.5 + 1\times0.35 + 4\times0.13 + 9\times0.02 = 1.05$ | M1 | $\Sigma x^2 p$ (at least 2 non-zero terms correct) |
| $\text{Var}(X) = \text{'their } 1.05\text{'} - 0.67^2 = 0.6011$ (awrt 0.6) | M1dep | for $(-0.67^2)$, provided $\text{Var}(X) > 0$ |
| | A1 | cao (No $n$ or $n-1$ divisors) — **Total: 3** |
---2 In a game of darts, a player throws three darts. Let $X$ represent the number of darts which hit the bull's-eye. The probability distribution of $X$ is shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = r )$ & 0.5 & 0.35 & $p$ & $q$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item (A) Show that $p + q = 0.15$.\\
(B) Given that the expectation of $X$ is 0.67 , show that $2 p + 3 q = 0.32$.\\
(C) Find the values of $p$ and $q$.
\item Find the variance of $X$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q2 [8]}}