S1 June 2011 Q8
8. A spinner is designed so that the score \(S\) is given by the following probability distribution.
| \(s\) | 0 | 1 | 2 | 4 | 5 |
|---|---|---|---|---|---|
| \(\mathrm{P}(S = s)\) | \(p\) | 0.25 | 0.25 | 0.20 | 0.20 |
Tom and Jess play a game with this spinner. The spinner is spun repeatedly and \(S\) counters are awarded on the outcome of each spin. If \(S\) is even then Tom receives the counters and if \(S\) is odd then Jess receives them. The first player to collect 10 or more counters is the winner.
| Scheme | Marks |
|---|---|
| \(1 = p + (0.25 + 0.25 + 0.2 + 0.2),\ \Rightarrow p = \underline{\tfrac{1}{10} \text{ or } 0.1}\) | M1, A1 |
| (2) |
Notes
M1 for clear attempt to use sum of probabilities = 1 (fractions or decimals) Ans only 2/2
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(S) = \tfrac{1}{4} + 2 \times \tfrac{1}{4} + 4 \times \tfrac{1}{5} + 5 \times \tfrac{1}{5}\), (or equiv. in decimals) = 2.55 | M1, A1 |
| (2) |
Notes
M1 for at least 2 correct terms (\(\neq 0\)) of the expression. 2.55 with no working scores M1A1
Any division by \(k\) (usually 5) in (b) or (c) or (d) scores M0
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(S^2) = \dfrac{1}{4} + \dfrac{2^2}{4} + \dfrac{4^2}{5} + \dfrac{5^2}{5}\) or 0.25 + 1 + 3.2 + 5 = 9.45 (*) | M1, A1cso |
| (2) |
Notes
M1 for at least 3 correct, non-zero terms of the expression seen, allow decimals.
A1cso for the full expression (with 9.45) seen. Must be cso but can ignore wrong \(p\).
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(S) = 9.45 - (\mathrm{E}(S))^2\), = 2.9475 or \(\dfrac{1179}{400}\) (accept awrt 2.95) | M1, A1 |
| (2) |
Notes
M1 for a correct expression (9.45 seen), can ft their E(\(S\)). May see \(\sum (x - \text{"}2.55\text{"})^2 \times \mathrm{P}(X = x)\)
A1 accept awrt 2.95 Answer only can score M1 for correct ft and A1 for awrt 2.95
Answer only in (e) and (f) is full marks, in (g) is no marks
| Scheme | Marks |
|---|---|
| P(5 and 5 ) = \(\left(\tfrac{1}{5}\right)^2\), = \(\tfrac{1}{25}\) or 0.04 | M1, A1 |
| (2) |
Notes
M1 for \(\left(\tfrac{1}{5}\right)^2\) Condone \(\mathrm{P}(5) \times \mathrm{P}(5) = 0.25 \times 0.25\). [Beware 0.4 is A0]
| Scheme | Marks |
|---|---|
| P(4, 4, 2) = \(\left(\tfrac{1}{5}\right)^2 \times \tfrac{1}{4} \times 3\) ( = 0.03 or \(\tfrac{3}{100}\)) | M1, M1 |
| P(4, 4, 4) = \(\left(\tfrac{1}{5}\right)^3\) ( = 0.008 or \(\tfrac{1}{125}\)) | B1 |
| P(Tom wins in 3 spins) = 0.038 | A1 |
| (4) |
Notes
1st M1 for \(\left(\tfrac{1}{5}\right)^2 \times \tfrac{1}{4}\) or 0.01 seen
2nd M1 for multiplying a \(p^2q\) probability by 3(\(p, q \in (0,1)\) ). B1 for \((0.2)^3\) or better seen
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(\overline{5} \cap 5 \cap 5) + \mathrm{P}(5 \cap \overline{5} \cap 5) = \tfrac{4}{5} \times \left(\tfrac{1}{5}\right)^2 \times 2\) = 0.064 or \(\tfrac{8}{125}\) | M1, M1, A1 |
| (3) | |
| (17 marks) |
Notes
1st M1 for \(\tfrac{4}{5} \times \left(\tfrac{1}{5}\right)^2\) or all cases considered and correct attempt at probabilities.
2nd M1 for multiplying a \(p^2(1 - p)\) probability by 2. Beware \((0.4)^3\) = 0.064 is M0M0A0