S1 June 2017 Q6
6. The score, \(X\), for a biased spinner is given by the probability distribution
| \(x\) | 0 | 3 | 6 |
|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(\dfrac{1}{12}\) | \(\dfrac{2}{3}\) | \(\dfrac{1}{4}\) |
Find
A biased coin has one face labelled 2 and the other face labelled 5
The score, \(Y\), when the coin is spun has
Sam plays a game with the spinner and the coin.
Each is spun once and Sam calculates his score, \(S\), as follows
if \(X = 0\) then \(S = Y^2\)
if \(X \neq 0\) then \(S = XY\)
Charlotte also plays the game with the spinner and the coin.
Each is spun once and Charlotte ignores the score on the coin and just uses \(X^2\) as her score.
Sam and Charlotte each play the game a large number of times.
| Scheme | Marks |
|---|---|
| \([\mathrm{E}(X)] = \left[0 \times \tfrac{1}{12}\right] + 3 \times \tfrac{2}{3} + 6 \times \tfrac{1}{4}\), \(= \tfrac{7}{2}\) or 3.5 | M1, A1 |
| (2) |
Notes
M1 for a fully correct expression (allow missing 0 term). Correct ans only is 2/2
| Scheme | Marks |
|---|---|
| \([\mathrm{E}(X^2)] = \left[0^2 \times \tfrac{1}{12}\right] + 3^2 \times \tfrac{2}{3} + 6^2 \times \tfrac{1}{4}\ \ (= 15)\) | M1 |
| \([\mathrm{Var}(X)] = \text{"}15\text{"} - \left(\text{"}\tfrac{7}{2}\text{"}\right)^2\) | M1 |
| \(= \tfrac{11}{4}\) or 2.75 | A1 |
| (3) |
Notes
1st M1 for a fully correct expression (allow missing 0 term) for E(\(X^2\)). Allow Var(\(X\)) label
2nd M1 for their E(\(X^2\)) – their E(\(X\))\(^2\)
| Scheme | Marks |
|---|---|
| \(5p + 2(1 - p) = 3\) or [ allow \(p + q = 1\) and \(5p + 2q = 3\) for M1] | M1A1 |
| So \(p = \tfrac{1}{3}\) (*) | A1 cso |
| (3) |
Notes
1st M1 for attempting a linear eq’n in \(p\)(or \(x\) etc). Must see = 3 and have 2 terms in \(p\), 1 correct
1st A1 for a fully correct equation for \(p\) or for solving their eqns leading to correct eqn in \(p\)
2nd A1 for \(p = \frac{1}{3}\) with M1 scored and no incorrect working seen.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(Y = 2) = \tfrac{2}{3}\) and \(\mathrm{P}(Y = 5) = \tfrac{1}{3}\) | B1 |
| (1) |
Notes
B1 for correct values for P(\(Y\) = 2) and P(\(Y\) = 5). Needn’t be in formal table but labelled.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(S = 30) = \mathrm{P}(X = 6 \text{ and } Y = 5)\) | M1 |
| \(= \tfrac{1}{4} \times \tfrac{1}{3} = \tfrac{1}{12}\) | A1cso |
| (2) |
Notes
M1 for \(6 \times 5 = 30\) or P(30) = P(6,5) or P(30) = P(6)\(\times\)P(5) or \(S = (XY =)\ 6 \times 5\) or \(X = 6\) and \(Y = 5\)
A1cso dep on M1 scored for with no incorrect working seen e.g. \(30 = \frac{1}{3} \times \frac{1}{4}\) is A0
| Scheme | Marks | ||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1A1A1 | ||||||||||||||
| (3) |
Notes
1st M1 for an attempt at prob. distribution with at least 3 correct (\(s\) and P(\(S = s\)))Exc’ \(s\)= 30
1st A1 for 6 correct \(s\) values 2nd A1 for a fully correct prob. distribution including \(s\) = 30
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(S) = \tfrac{1}{36}[4 \times 2 + 6 \times 16 + 12 \times 6 + 15 \times 8 + 25 \times 1 + 30 \times 3]\) | M1 |
| \(= 11\tfrac{5}{12}\) or \(\tfrac{137}{12}\) or \(11.41\dot{6}\) | A1 |
| (2) |
Notes
M1 for attempting E(\(S\)) using their values. Must see …3 products (correct ft) decimals to 3sf
A1 for \(11\frac{5}{12}\) or \(\frac{137}{12}\) or any exact equivalent. (Correct ans. only 2/2, awrt 11.4 only M1A0)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^2) = 15\) and \(\mathrm{E}(S) = 11.416\ldots\) or \(\mathrm{E}(X^2) \gt \mathrm{E}(S)\) | B1ft |
| … so Charlotte has the higher total score | dB1ft |
| (2) | |
| (18 marks) |
Notes
1st B1 for correct comparison of their E(\(S\)) and E(\(X^2\)) labelled in (b) or (h) [expressions or values]
2nd d B1 dependent on a correct comparison of their values for choosing correct player.