| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Three or more stages |
| Difficulty | Moderate -0.3 This is a standard three-stage tree diagram problem with conditional probability. While it requires careful tracking of disc movements between bags and multiple probability calculations, the techniques are routine for S1 students. The tree diagram structure is provided, making it primarily an exercise in applying basic probability rules (multiplication along branches, addition across outcomes) rather than requiring problem-solving insight. The variance calculation adds some computational work but remains standard bookwork. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables5.02b Expectation and variance: discrete random variables |
| \(r\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( R = r )\) | \(\frac { 1 } { 10 }\) | \(k\) | \(\frac { 9 } { 20 }\) | \(\frac { 1 } { 5 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a = \frac{1}{8}\), \(b = \frac{1}{8}\); \(c = \frac{1}{4}\), \(d = \frac{3}{4}\); \(e = \frac{3}{4}\), \(f = \frac{1}{4}\) | B1 B1B1 B1 | 4 |
| (ii) \(\frac{1}{2X} \times \frac{1}{3} \times \frac{1}{2} + \frac{2}{5X} \times \frac{1}{X} \times \frac{1}{4} + \frac{7}{5X} \times \frac{1}{X} \times \frac{1}{4}\) | M2 | M1: one correct product; NOT \(\frac{1}{2X}C_2\times C_1\); M2 needs +) fit their values for M mks only |
| \(= \frac{29}{20}\) (AG) with no errors seen | A1 | 3 |
| (iii) \(\frac{1}{10} + \frac{9}{20} + k \times \frac{1}{5} = 1\) oe or \(\frac{1}{x}x^0 \times \frac{1}{4} + \frac{1}{x}x^{1/2} \times \frac{1}{6} + \frac{1}{x}x^{1/3} \times \frac{1}{2}\); \(k = \frac{1}{4}\) oe | M1 A1 | 2 |
| (iv) \(\Sigma xp(x)\) | M1 A1 | |
| \(= 1\frac{3}{4}\) oe | ||
| \(\Sigma x^2 p(x) = \left[3\frac{13}{20}\right]\) | M1 M1ind | Allow omit 1st term only. Not ISW, eg ÷ 4; Subtract (their \(\mu\))²; if result +ve |
| \(\Sigma x^2p(x) - (\text{their } \mu)^2 = \frac{63}{80}\) or \(0.788\) (3 sfs) | A1 | 5 |
(i) $a = \frac{1}{8}$, $b = \frac{1}{8}$; $c = \frac{1}{4}$, $d = \frac{3}{4}$; $e = \frac{3}{4}$, $f = \frac{1}{4}$ | B1 B1B1 B1 | 4 | Or: B1 $\{$ ie: $a, b$ $\}$: B1; another pair: B1B1; third pair: B1
(ii) $\frac{1}{2X} \times \frac{1}{3} \times \frac{1}{2} + \frac{2}{5X} \times \frac{1}{X} \times \frac{1}{4} + \frac{7}{5X} \times \frac{1}{X} \times \frac{1}{4}$ | M2 | M1: one correct product; NOT $\frac{1}{2X}C_2\times C_1$; M2 needs +) fit their values for M mks only
$= \frac{29}{20}$ (AG) with no errors seen | A1 | 3 |
(iii) $\frac{1}{10} + \frac{9}{20} + k \times \frac{1}{5} = 1$ oe or $\frac{1}{x}x^0 \times \frac{1}{4} + \frac{1}{x}x^{1/2} \times \frac{1}{6} + \frac{1}{x}x^{1/3} \times \frac{1}{2}$; $k = \frac{1}{4}$ oe | M1 A1 | 2 | fit their values for M mk only
(iv) $\Sigma xp(x)$ | M1 A1 | | Allow omit 1st term only. Not ISW, eg ÷ 4 cao
$= 1\frac{3}{4}$ oe |
$\Sigma x^2 p(x) = \left[3\frac{13}{20}\right]$ | M1 M1ind | Allow omit 1st term only. Not ISW, eg ÷ 4; Subtract (their $\mu$)²; if result +ve
$\Sigma x^2p(x) - (\text{their } \mu)^2 = \frac{63}{80}$ or $0.788$ (3 sfs) | A1 | 5 | Follow their $k$ for M mks only; $\Sigma(x - \mu)^2p(x)$: Single consistent pair: M1; Rest correct: M16 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]{5faf0d93-4037-4958-8665-1008477a79de-5_863_986_559_612}\\
(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 2005 Q6 [14]}}