S1 June 2018 Q4
4. A bag contains 64 coloured beads. There are \(r\) red beads, \(y\) yellow beads and 1 green bead and \(r + y + 1 = 64\)
Two beads are selected at random, one at a time without replacement.
The probability that both of the beads are red is \(\dfrac{5}{84}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(G_1) + \mathrm{P}(R_1 \cap G_2) + \mathrm{P}(Y_1 \cap G_2)\) or P(\(GY\)) + P(\(GR\)) + P(\(RG\)) + P(\(YG\)) (o.e.) | M1 |
| \(= \dfrac{1}{64} + \dfrac{r}{64} \times \dfrac{1}{63} + \dfrac{y}{64} \times \dfrac{1}{63} = \dfrac{1}{64} + \dfrac{r + y}{64 \times 63}\) or \(2 \times \dfrac{r + y}{64 \times 63}\) | A1 |
| \(= \dfrac{1}{64} + \dfrac{63}{64 \times 63}\) or \(\dfrac{2 \times 63}{64 \times 63}\) or \(\dfrac{1}{64} + \dfrac{1}{64}\) or | M1 |
| \(= \dfrac{1}{32}\) or 0.03125 | A1 |
| (4) |
Notes
1st M1 for at least 2 correct cases. May be in symbols or probs. May be in tree diagram
Use of \(r = 16\) or \(y = 47\) can score maximum of 1st M1 then A0M0A0
1st A1 for all cases and their assosciated probs added
2nd M1 for combining probabilities and using \(r + y = 63\)
2nd A1 for \(\frac{1}{32}\) or an exact equivalent (correct answer only 4/4)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(R_1 \cap R_2) = \dfrac{r}{64} \times \dfrac{r - 1}{63} = \dfrac{5}{84}\) | M1A1 |
| \(r(r - 1) = 5 \times 64 \times 63 \div 84 = 240\) hence \(r^2 - r - 240 = 0\) or \(r^2 - r = 240\) (*) | A1cso |
| (3) |
Notes
M1 for \(\frac{r}{64} \times \mathrm{g}(r) = \ldots\) where \(\mathrm{g}(r)\) is any linear function of \(r\)
1st A1 for any correct equation in \(r\)
2nd A1cso for correctly simplifying to the given equation with no incorrect working seen.
There should be at least 1 intermediate step seen
| Scheme | Marks |
|---|---|
| \(r^2 - r - 240 = (r - 16)(r + 15)\{= 0\}\) or \(16^2 - 16 - 240 = 256 - 256\) or \(\frac{16}{64} \times \frac{15}{63} = \frac{5}{84}\) | M1 |
| so \(r = 16\) and rejecting \(-15\) (*) | A1cso |
| (2) |
Notes
M1 for correct factors or completing square or use of formula or substitution
A1cso for concluding \(r = 16\) and rejecting – 15 (e.g. crossing out etc)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(\geqslant 1 \text{ red}) =\) P(\(RG\)) + P(\(GR\)) + P(\(RY\)) + P(\(YR\)) + P(\(RR\)) or \(\frac{2}{252} + \frac{2y}{252} + \frac{15}{252}\) (o.e.) or \(\mathrm{P}(R_1) + \mathrm{P}(R_1^{\prime} \cap R_2)\) or \(\dfrac{16}{64} + \dfrac{48}{64} \times \dfrac{16}{63}\) or \(1 - \dfrac{48}{64} \times \dfrac{47}{63}\), \(= \dfrac{37}{84}\) | M1, A1 |
| Require: \(\dfrac{\mathrm{P}(R_1 \cap R_2)}{\mathrm{P}(\text{at least one red})} = \dfrac{\frac{5}{84}}{\text{"}\frac{37}{84}\text{"}}\) \(,= \dfrac{5}{37}\) or \(0.\dot{1}3\dot{5}\) | M1, A1 |
| (4) | |
| (13 marks) |
Notes
1st M1 for a correct expression for at least one red. May be in symbols or probs. or in a tree
1st A1 for \(\frac{37}{84}\) (o.e.) as a single fraction or awrt 0.440 [May be implied by correct answer]
2nd M1 for a ratio of probabilities (denom may be in symbols) with numerator of \(\frac{5}{84}\) (o.e.)
2nd A1 for \(\frac{5}{37}\) or an exact equivalent