Foundation January 2019 Paper 1 Q15
15 Here is a biased 5-sided spinner.
Kenny spins the spinner once.
The table gives the probabilities that the spinner lands on red or on blue or on green.
| Colour | red | blue | green | brown | yellow |
|---|---|---|---|---|---|
| Probability | 0.15 | 0.26 | 0.33 |
(a) Work out the probability that the spinner lands on red or blue. (1)
When the spinner is spun once, the probability that the spinner lands on brown is 0.06 more than the probability that the spinner lands on yellow.
Jenine spins the spinner 150 times.
(b) Work out an estimate for the number of times the spinner lands on yellow. (4)
| Scheme | Marks |
|---|---|
| 0.15 + 0.26 Answer: 0.41 oe | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| 1 − (0.15 + 0.26 + 0.33) or 1 − 0.74 (= 0.26) | M1 |
| (P(yellow) =) \(\dfrac{\text{``}{0.26}\text{''} - 0.06}{2}\) or 0.1 | M1 |
| 150 × “0.1” | M1 |
| 15 | A1 |
| (4) | |
| (5 marks) |
Notes
M1: can be implied by two values where P(brown) + P(yellow) = 0.26 (may be seen in table)
M1: for a complete method to find P(yellow)
M1: independent mark
Award for 150 × \(p\) where 0 < \(p\) < 1
A1: NB: An answer of \(\dfrac{15}{150}\) scores M3 A0