S1 June 2010 Q2
2. An experiment consists of selecting a ball from a bag and spinning a coin. The bag contains 5 red balls and 7 blue balls. A ball is selected at random from the bag, its colour is noted and then the ball is returned to the bag.
When a red ball is selected, a biased coin with probability \(\tfrac{2}{3}\) of landing heads is spun.
When a blue ball is selected a fair coin is spun.
Shivani selects a ball and spins the appropriate coin.
Given that Tom selected a ball at random and obtained a head when he spun the appropriate coin,
Shivani and Tom each repeat this experiment.
| Scheme | Marks |
|---|---|
 | |
| P(\(R\)) and P(\(B\)) | B1 |
| 2nd set of probabilities | B1 |
| (2) |
Notes
1st B1 for the probabilities on the first 2 branches. Accept \(0.41\dot{6}\) and \(0.58\dot{3}\)
2nd B1 for probabilities on the second set of branches. Accept \(0.\dot{6}\), \(0.\dot{3}\), 0.5 and \(\dfrac{1.5}{3}\)
Allow exact decimal equivalents using clear recurring notation if required.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(H) = \dfrac{5}{12} \times \dfrac{2}{3} + \dfrac{7}{12} \times \dfrac{1}{2},\ = \dfrac{41}{72}\) or awrt 0.569 | M1 A1 |
| (2) |
Notes
M1 for an expression for P(\(H\)) that follows through their sum of two products of probabilities from their tree diagram
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(R|H) = \dfrac{\frac{5}{12} \times \frac{2}{3}}{\text{"}\frac{41}{72}\text{"}},\ = \dfrac{20}{41}\) or awrt 0.488 | M1 A1ft A1 |
| (3) |
Notes
Formula seen: M1 for \(\dfrac{\mathrm{P}(R \cap H)}{\mathrm{P}(H)}\) with denominator their (b) substituted e.g. \(\dfrac{\mathrm{P}(R \cap H)}{\mathrm{P}(H)} = \dfrac{\frac{5}{12}}{(\text{their (b)})}\) award M1.
Formula not seen: M1 for \(\dfrac{\text{probability} \times \text{probability}}{\text{their } b}\) but M0 if fraction repeated e.g. \(\dfrac{\frac{5}{12} \times \frac{2}{3}}{\frac{2}{3}}\).
1st A1ft for a fully correct expression or correct follow through
2nd A1 for \(\tfrac{20}{41}\) o.e.
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{5}{12}\right)^2 + \left(\dfrac{7}{12}\right)^2\) | M1 A1ft |
| \(= \dfrac{25}{144} + \dfrac{49}{144} = \dfrac{74}{144}\) or \(\dfrac{37}{72}\) or awrt 0.514 | A1 |
| (3) | |
| (10 marks) |
Notes
M1 for \(\left(\dfrac{5}{12}\right)^2\) or \(\left(\dfrac{7}{12}\right)^2\) can follow through their equivalent values from tree diagram
1st A1 for both values correct or follow through from their original tree and +
2nd A1 for a correct answer
Special Case \(\dfrac{5}{12} \times \dfrac{4}{11}\) or \(\dfrac{7}{12} \times \dfrac{6}{11}\) seen award M1A0A0