| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct distribution then calculate probability |
| Difficulty | Standard +0.3 This is a multi-part question involving conditional probability with a tree diagram structure. Part (i) requires calculating straightforward conditional probabilities from bag compositions. Parts (ii)-(iv) involve constructing and working with a discrete probability distribution using standard formulas. While it has multiple steps and requires careful tracking of bag contents, the techniques are all routine S1 material with no novel problem-solving required. Slightly easier than average due to the scaffolded structure and given tree diagram. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables5.01a Permutations and combinations: evaluate probabilities |
| \(r\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( R = r )\) | \(\frac { 1 } { 10 }\) | \(k\) | \(\frac { 9 } { 20 }\) | \(\frac { 1 } { 5 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | \(\frac{1}{5} \times \frac{1}{4} \times 2\) or \(\frac{2}{5} \times \frac{1}{4}\) alone oe; = \(\frac{1}{10}\) or 0.1 oe | M1, A1 |
| 6(ii) | \(\frac{2}{5} \times \frac{1}{4} \times \frac{2}{3}\) or \(\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times 2\) oe; or \(\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times 3\); or 1– (\(\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3}\) + \(\frac{2}{5} \times \frac{3}{4} \times \frac{1}{3} \times 3\)); = \(\frac{4}{5}\) or 0.8 oe | M1, A1 |
**6(i)** | $\frac{1}{5} \times \frac{1}{4} \times 2$ or $\frac{2}{5} \times \frac{1}{4}$ alone oe; = $\frac{1}{10}$ or 0.1 oe | M1, A1 | Allow M1 for $\frac{1}{5} \times \frac{1}{4}$, but NOT other methods leading to $\frac{1}{20}$ and NOT $\frac{1}{20}$ with no wking |
**6(ii)** | $\frac{2}{5} \times \frac{1}{4} \times \frac{2}{3}$ or $\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times 2$ oe; or $\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times 3$; or 1– ($\frac{3}{5} \times \frac{2}{4} \times \frac{1}{3}$ + $\frac{2}{5} \times \frac{3}{4} \times \frac{1}{3} \times 3$); = $\frac{4}{5}$ or 0.8 oe | M1, A1 | Only if using complement (ie 1–P(OV or 2V)); See comment before 6(i) |6 Two bags contain coloured discs. At first, bag $P$ contains 2 red discs and 2 green discs, and bag $Q$ contains 3 red discs and 1 green disc. A disc is chosen at random from bag $P$, its colour is noted and it is placed in bag $Q$. A disc is then chosen at random from bag $Q$, its colour is noted and it is placed in bag $P$. A disc is then chosen at random from bag $P$.
The tree diagram shows the different combinations of three coloured discs chosen.\\
\includegraphics[max width=\textwidth, alt={}, center]{11316ea6-3999-4003-b77d-bee8b547c1da-05_858_980_573_585}\\
(i) Write down the values of $a , b , c , d , e$ and $f$.
The total number of red discs chosen, out of 3, is denoted by $R$. The table shows the probability distribution of $R$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( R = r )$ & $\frac { 1 } { 10 }$ & $k$ & $\frac { 9 } { 20 }$ & $\frac { 1 } { 5 }$ \\
\hline
\end{tabular}
\end{center}
(ii) Show how to obtain the value $\mathrm { P } ( R = 2 ) = \frac { 9 } { 20 }$.\\
(iii) Find the value of $k$.\\
(iv) Calculate the mean and variance of $R$.
\hfill \mbox{\textit{OCR S1 Q6 [13]}}