4MA0 Higher June 2015 Paper 4R Q16
16 When a fair dice is thrown the probability of scoring 6 is \(\dfrac{1}{6}\)
Arun throws four fair dice.
Work out the probability that he scores 6 with at least one of the four dice.
(3)
| Scheme | Marks |
|---|---|
\(4\left(\dfrac{1}{6}\right)\left(\dfrac{5}{6}\right)^3 + 6\left(\dfrac{1}{6}\right)^2\left(\dfrac{5}{6}\right)^2 + 4\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right) + \left(\dfrac{1}{6}\right)^4\) or \(\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right) \times \left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right) \times \left(\dfrac{5}{6}\right) \times \left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right) \times \left(\dfrac{5}{6}\right) \times \left(\dfrac{5}{6}\right) \times \left(\dfrac{1}{6}\right)\) or \(1 - \left(\dfrac{5}{6}\right)^4\) | M2 |
| \(\dfrac{671}{1296}\) | A1 |
| (3) |
Notes
M2: Accept use of 0.16(666..) in place of \(\dfrac{1}{6}\) and 0.83(3333...) in place of \(\dfrac{5}{6}\) rounded or truncated to at least 2 decimal places.
M1 for \(\left(\dfrac{1}{6}\right) \times \left(\dfrac{5}{6}\right)^3\) or \(\left(\dfrac{1}{6}\right)^2 \times \left(\dfrac{5}{6}\right)^2\) or \(\left(\dfrac{1}{6}\right)^3 \times \left(\dfrac{5}{6}\right)\) or \(\left(\dfrac{1}{6}\right)^4\) or P(no sixes) = \(\left(\dfrac{5}{6}\right)^4\)
A1: Accept awrt 0.518