4MA0 Higher June 2015 Paper 3R Q17
17 Two bags contain discs.
Bag A contains 12 discs.
5 of the discs are red, 6 are blue and 1 is white.
Bag B contains 25 discs.
\(n\) of the discs are red and the rest are blue.
James takes at random a disc from Bag A.
Lucy takes at random a disc from Bag B.
Given that the probability that James and Lucy both take a red disc is \(\dfrac{2}{15}\)
(i) find the value of \(n\), the number of red discs in Bag B.
(ii) Hence calculate the probability that James and Lucy take discs of different colours.
(5)
| Scheme | Marks |
|---|---|
\(\dfrac{5}{12} \times \dfrac{n}{25}\left(= \dfrac{2}{15}\right)\) oe or \(\dfrac{2}{15} \div \dfrac{5}{12}\) oe | M1 |
| 8 | A1 |
| (2) |
Notes
A1: SC: B1 for an answer of \(\dfrac{8}{25}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{6}{12} \times \dfrac{25 - \text{"}8\text{"}}{25}\left(= \dfrac{17}{50}\right)\) | M1 |
| \(1 - \text{"}\dfrac{17}{50}\text{"} - \dfrac{2}{15}\) | M1 |
| \(\dfrac{79}{150}\) oe | A1 |
| (3) |
Notes
M1: ft from (a)
for one correct branch from (R, B) or (B, R) or (W, R) or (W, B)
\(\dfrac{5}{12} \times \dfrac{25 - \text{"}8\text{"}}{25}\left(= \dfrac{17}{60}\right)\) or \(\dfrac{6}{12} \times \dfrac{\text{"}8\text{"}}{25}\left(= \dfrac{4}{25}\right)\) or \(\dfrac{1}{12} \times \dfrac{\text{"}8\text{"}}{25}\left(= \dfrac{2}{75}\right)\) or \(\dfrac{1}{12} \times \dfrac{25 - \text{"}8\text{"}}{25}\left(= \dfrac{17}{300}\right)\)
NB: \(\dfrac{7}{12} \times \dfrac{\text{"}8\text{"}}{25}\) implies BR + WR; \(\dfrac{6}{12} \times \dfrac{25 - \text{"}8\text{"}}{25}\) implies RB and WB
\(\dfrac{1}{12}\) implies WB + WR
M1: ft from (a) for all products with the intention to add
A1: cao accept 0.527 or 0.526…. or 52.7% or 52.6..%