A2 June 2025 Q6
6. A biased coin and a fair coin are thrown repeatedly.
The probability that the biased coin shows heads is \(\dfrac{1}{3}\)
Let \(B\) represent the number of throws of the biased coin until it shows heads for the first time.
Let \(F\) represent the number of throws of the fair coin until it shows heads for the first time.
Each day Chris invites his friend Shivani to play a game with these coins.
One player’s score will be \(X\) and the other player’s score will be \(Y\), where
Chris and Shivani will play a large number of games and the winner will be the player with the higher total score.
Before their first game, Chris allows Shivani to choose whether her score for every game will be \(X\) or her score for every game will be \(Y\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(B \sim \mathrm{Geo}\left(\dfrac{1}{3}\right)\) | M1 | 3.3 |
| \(\mathrm{P}(B = 3) = \left(\dfrac{2}{3}\right)^2 \times \left(\dfrac{1}{3}\right) = \dfrac{4}{27}\) awrt 0.148 | A1 | 1.1b |
| (2) | ||
| (ii) \(F \sim \mathrm{Geo}\left(\dfrac{1}{2}\right)\) and use of \(\mathrm{P}(F \leqslant n) = 1 - (1 - p)^n\) | M1 | 2.1 |
| \(\mathrm{P}(2 \leqslant F \leqslant 8) = \mathrm{P}(F \leqslant 8) - \mathrm{P}(F \leqslant 1)\) | M1 | 1.1b |
| \(\mathrm{P}(2 \leqslant F \leqslant 8) = 1 - \left(\dfrac{1}{2}\right)^8 - \left(1 - \dfrac{1}{2}\right) = \dfrac{1}{2} - \dfrac{1}{256} = \dfrac{127}{256}\) awrt 0.496 | A1 | 1.1b |
| (3) | ||
| (iii) \(\mathrm{P}[(F = 3) \cup (B = 3)] = \mathrm{P}(F = 3) + \mathrm{P}(B = 3) - \mathrm{P}(F = 3) \times \mathrm{P}(B = 3)\) or \(\mathrm{P}[(F = 3) \cup (B = 3)] = 1 - \mathrm{P}(F \neq 3) \times \mathrm{P}(B \neq 3)\) or \(\mathrm{P}[(F = 3) \cup (B = 3)] = \mathrm{P}(B = 3) + \mathrm{P}(F = 3) \times \mathrm{P}(B \neq 3)\) | M1 | 2.1 |
| \(\text{‘}\dfrac{4}{27}\text{’} + \left[\left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right] - \text{‘}\dfrac{4}{27}\text{’} \times \left[\left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right]\) or \(1 - \left[1 - \text{‘}\dfrac{4}{27}\text{’}\right] \times \left[1 - \left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right]\) | M1 | 1.1b |
| \(\text{‘}\dfrac{4}{27}\text{’} + \dfrac{1}{8} - \text{‘}\dfrac{4}{27}\text{’} \times \dfrac{1}{8} = \dfrac{55}{216}\) or \(1 - \left[1 - \text{‘}\dfrac{4}{27}\text{’}\right] \times \dfrac{7}{8} = \dfrac{55}{216}\) awrt 0.255 | A1 | 1.1b |
| (3) |
Notes
(i) M1: For sight or use of the correct geometric model
A1: \(\dfrac{4}{27}\) or awrt 0.148 (correct answer implies full marks)
(ii) M1: For selecting the correct geometric model and use of cumulative probability formula
Implied by \(\dfrac{1}{256}\)
M1: Sight or use of \(\mathrm{P}(F \leqslant 8) - \mathrm{P}(F \leqslant 1)\)
A1: \(\dfrac{127}{256}\) or awrt 0.496 (correct answer implies full marks)
(iii) M1: Sight or use of
\(\mathrm{P}[(F = 3) \cup (B = 3)] = \mathrm{P}(F = 3) + \mathrm{P}(B = 3) - \mathrm{P}(F = 3) \times \mathrm{P}(B = 3)\)
or
\(\mathrm{P}[(F = 3) \cup (B = 3)] = 1 - \mathrm{P}(F \neq 3) \times \mathrm{P}(B \neq 3)\)
or
\(\mathrm{P}[(F = 3) \cup (B = 3)] = \mathrm{P}(B = 3) + \mathrm{P}(F = 3) \times \mathrm{P}(B \neq 3)\) (corrected from the printed mark scheme: the notes print this as \(\mathrm{P}(B = 3) - \mathrm{P}(F = 3) \times \mathrm{P}(B \neq 3)\); the scheme above has \(+\))
M1: \(\text{‘}\dfrac{4}{27}\text{’} + \left[\left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right] - \text{‘}\dfrac{4}{27}\text{’} \times \left[\left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right]\)
or
\(1 - \left[1 - \text{‘}\dfrac{4}{27}\text{’}\right] \times \left[1 - \left(\dfrac{1}{2}\right)^2 \times \left(\dfrac{1}{2}\right)\right]\)
ft their \(\mathrm{P}(B = 3)\) from (a)(i)
A1: \(\dfrac{55}{216}\) or awrt 0.255 (correct answer implies full marks)
| Scheme | Marks | AO |
|---|---|---|
| Require \(\mathrm{E}(4F)\) and \(\mathrm{E}\left(\dfrac{B^2}{2}\right)\) | M1 | 3.1a |
| \(\mathrm{E}(X) = \mathrm{E}(4F) = 4\mathrm{E}(F) = 4 \times 2 = 8\) \(\mathrm{E}(X) = 8\) | B1 | 2.1 |
| \(\mathrm{E}(Y) = \mathrm{E}\left(\dfrac{B^2}{2}\right) = \dfrac{1}{2}\mathrm{E}(B^2) = \dfrac{1}{2}\left(\mathrm{Var}(B) + [\mathrm{E}(B)]^2\right)\) | M1 | 1.1b |
| \(\mathrm{E}\left(\dfrac{B^2}{2}\right) = \dfrac{1}{2}\left(\dfrac{1 - \frac{1}{3}}{\left(\frac{1}{3}\right)^2} + (3)^2\right) = \dfrac{1}{2}(6 + 9)\) \(\mathrm{E}(Y) = 7.5\) | A1 | 1.1b |
| Shivani should choose \(X\) as it has a higher expected score | A1ft | 3.2a |
| (5) | ||
| (13 marks) |
Notes
M1: For attempting to calculate or consider expected values/means of \(X\) and \(Y\)
Implied by sight of 8 and 4.5
B1: for \(\mathrm{E}(X) = 8\)
M1: Using \(\mathrm{E}(B^2) = \mathrm{Var}(B) + [\mathrm{E}(B)]^2\)
or
\(\mathrm{E}(B^2) = \mathrm{G}'_B(1) + \mathrm{G}''_B(1)\) with a correct PGF given for \(B\)
\(\mathrm{E}(B^2) = 15\) is M1
A1: for \(\mathrm{E}(Y) = 7.5\) or equivalent
A1ft: for interpreting the outcome of their calculations in terms of the problem.
Must clearly choose \(X\) or \(Y\) and refer to its expectation/average being higher
Allow reference to expectation as a comparison e.g. \(8 \gt 7.5\)
ft their \(\mathrm{E}(X)\) and \(\mathrm{E}(Y)\) provided at least one of \(\mathrm{E}(X)\) or \(\mathrm{E}(Y)\) correct
dependent on 1st M1 being awarded
e.g. candidates can score M1B1M0A0A1 or M1B0M1A1A1 or M1B1M1A0A1
NB: Accept expectations given as totals. E.g. \(8n\) and \(7.5n\), but not \(7.5n^2\) etc