AS June 2022 Q4
4. The discrete random variable \(X\) has the following probability distribution
| \(x\) | 0 | 2 | 3 | 6 |
|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(p\) | 0.25 | \(q\) | 0.4 |
Given that \(\mathrm{Var}(X) = 3.66\)
In a game, the score is given by the discrete random variable \(X\)
Given that games are independent,
A round consists of 4 games plus 2 bonus games. The bonus games are only played if after the 4th game has been played the total score is exactly 20
A prize of £10 is awarded if 6 games are played in a round and the total score for the round is at least 27
Bobby plays 3 rounds.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathrm{E}(X) = [0 \times p] + (2 \times 0.25) + 3q + (6 \times 0.4)\ [= 2.9 + 3q]\) | B1 | 1.1b |
| (ii) \(\mathrm{E}(X^2) = [0 \times p] + (2^2 \times 0.25) + 3^2 q + (6^2 \times 0.4)\ [= 15.4 + 9q]\) | B1 | 1.1b |
| (2) |
Notes
(i) B1: Correct expression for \(\mathrm{E}(X)\) need not be simplified
(ii) B1: Correct expression for \(\mathrm{E}(X^2)\) need not be simplified
| Scheme | Marks | AO |
|---|---|---|
| \((\text{“}15.4 + 9q\text{”}) - (\text{“}2.9 + 3q\text{”})^2 = 3.66\) | M1 | 1.1b |
| \(9q^2 + 8.4q - 3.33 = 0 \Rightarrow q = 0.3\) and \(-\tfrac{37}{30}\) | M1 | 1.1b |
| \(q = 0.3\)* since \(q\) cannot be negative | A1cso* | 2.4 |
| SC \((\text{“}15.4 + 9 \times 0.3\text{”}) - (\text{“}2.9 + 3 \times 0.3\text{”})^2\) can get M1M0A0 | ||
| (3) |
Notes
M1: Using "their \(\mathrm{E}(X^2)\)" – "their \((\mathrm{E}(X))^2\)" = 3.66
M1: Rearranging to get a correct 3 term quadratic (condone missing = 0) leading to 0.3 and – 37/30(awrt –1.23) or \((10q - 3)(30q + 37)\)
A1cso*: cso with a comment why –37/30 is eliminated. Minimum required is \(q \gt 0\) or they say it is impossible.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(x_1 + x_2 + x_3 + x_4 = 20) = \mathrm{P}(6,6,6,2 \text{ or } 6,6,2,6 \text{ or } 6,2,6,6 \text{ or } 2,6,6,6)\) | M1 | 1.1b |
| \(= 4 \times 0.4^3 \times 0.25\) | M1 | 1.1b |
| \(= 0.064\) oe | A1 | 1.1b |
| (3) |
Notes
M1: Realising that combination is 6662. Any order. Implied by \(0.4^3 \times 0.25\)
M1: Correct calculation
A1: 0.064 oe only eg 8/125
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(x_5 + x_6 \geqslant 7) = \mathrm{P}(6,6 \text{ or } 6,3 \text{ or } 6,2)\) | M1 | 3.1a |
| \(= (0.4^2) + 2 \times (0.4 \times 0.3) + 2 \times 0.4 \times 0.25\ [= 0.6]\) | M1 | 1.1b |
| \(\mathrm{P}(\text{score} \geqslant 27) = \text{“}0.064\text{”} \times \text{“}0.6\text{”}\ [= 24/625 = 0.0384]\) | M1 | 1.1b |
| \(Y \sim \mathrm{B}(3, \text{“}0.0384\text{”})\) | dM1 | 3.3 |
| \(\mathrm{P}(Y \geqslant 1) = 1 - \mathrm{P}(Y = 0)\) | M1 | 1.1b |
| \(= 0.1108\ldots\) | A1cso | 1.1b |
| (6) | ||
| (14 marks) |
Notes
M1: Realising all the different combinations 7 or more can be scored from 2 games. (no need for arrangements) Implied by \((0.4^2)\) and \((0.4 \times 0.3)\) and \((0.4 \times 0.25)\)
M1: Fully correct method.
M1: For multiplying "their (c)" with "their \(\mathrm{P}(x_5 + x_6 \geqslant 7)\)" providing at least 2 combinations are used to find \(\mathrm{P}(x_5 + x_6 \geqslant 7)\)
dM1: Dependent on 3rd M1 being awarded for using or writing \(\mathrm{B}(3, \text{“their } \mathrm{P}(x_1 + x_2 + x_3 + x_4 + x_5 + x_6 \geqslant 27)\text{”})\) \((1 - \text{“}0.0384\text{”})^3\) or
M1: For writing or using \(1 - \mathrm{P}(Y = 0)\) eg \(1 - (1 - \text{“}0.0384\text{”})^3\)
A1cso: awrt 0.111 from correct working
NB (d) 1st 3 marks (corrected from the printed mark scheme: this note is headed “NB (b)” there, but it is about the first three marks of part (d))
| Fully correct method \(\text{“}0.064\text{”} \times (0.4^2) + 0.064 \times 2 \times (0.4 \times 0.3) + 0.064 \times 2 \times (0.4 \times 0.25)\) | M1M1M1 |
| All 3 but no arrangements ie \(\text{“}0.064\text{”} \times (0.4^2) + 0.064 \times (0.4 \times 0.3) + 0.064 \times (0.4 \times 0.25)\) | M1M0M1 |
| At least 2 combinations used for > 7 eg \(0.064 \times (0.4 \times 0.3) + 0.064 \times (0.4^2)\) or \(2 \times (0.4 \times 0.3)\) | M0M0M1 |