4MA0 Higher January 2017 Paper 3 Q24
24 The diagram shows three boxes containing beads.
Each box contains 3 black beads and 6 white beads.
Tim takes at random a bead from box A and puts it into box B.
He then takes at random a bead from box B and puts it into box C.
Finally, he takes at random a bead from box C and puts it into box A.
Calculate the probability that there are still 3 black beads and 6 white beads in each of the three boxes.
(3)
| Scheme | Marks |
|---|---|
\(\dfrac{6}{9} \times \dfrac{7}{10} \times \dfrac{7}{10}\left(= \dfrac{49}{150}\right)\) oe or \(\dfrac{3}{9} \times \dfrac{4}{10} \times \dfrac{4}{10}\left(= \dfrac{4}{75}\right)\) oe OR \(\dfrac{6}{9} \times \dfrac{7}{10} \times a\) and \(\dfrac{3}{9} \times \dfrac{4}{10} \times b\) \(a\) and \(b\) must both be a single fraction where \(0 \lt a, b \lt 1\) and \(a \ne \dfrac{7}{10},\; b \ne \dfrac{4}{10}\) | M1 |
| \(\dfrac{6}{9} \times \dfrac{7}{10} \times \dfrac{7}{10}\) oe and \(\dfrac{3}{9} \times \dfrac{4}{10} \times \dfrac{4}{10}\) oe | M1 |
| \(\dfrac{19}{50}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: Both products correct
(addition not needed)
A1: oe E.g. \(\dfrac{342}{900}\)