4400 Higher June 2006 Paper 3 Q14
14 Here is a biased spinner.
When the pointer is spun, the score is 1 or 2 or 3 or 4
The probability that the score is 1 is 0.3
The probability that the score is 2 is 0.6
Hajra spins the pointer once.
(a) Work out the probability that
(i) the score is 1 or 2
(ii) the score is 3 or 4 (3)
Nassim spins the pointer twice.
(b) Work out the probability that
(i) the score is 1 both times,
(ii) the score is 2 exactly once. (5)
| Scheme | Marks |
|---|---|
| (i) 0.3 + 0.6 | M1 |
| 0.9 | A1 |
| (ii) 0.1 | B1 |
| (3) |
Notes
B1: ft 1 − “0.9”
| Scheme | Marks |
|---|---|
| (i) 0.3 × 0.3 | M1 |
| 0.09 | A1 |
| (ii) 0.6×0.4 or 0.6×0.3 + 0.6×0.1 or 0.24 or 0.6×0.3×2 or 0.36 or 0.6×0.1×2 or 0.12 | M1 |
| “0.24” × 2 oe | M1 |
| 0.48 | A1 |
| (5) | |
| (8 marks) |
Notes
M1: dep on previous M1
A1: ft from “0.1”
The assumption that P(3) = P(4) = 0.5 makes the method incorrect and 0.48 cannot gain full marks but 0.6 × 0.3 × 2 still scores M1.
Alternative for (b)(ii)
| \(1 - (0.6 \times 0.6 + 0.4 \times 0.4)\) or \(1 - (0.36 + 0.16)\) | M2 |
| 0.48 | A1 |