Foundation June 2025 Paper 1R Q19
19 Here is a biased 5-sided spinner.
When the spinner is spun, it can land on blue or on green or on purple or on red or on yellow.
The table gives information about the probability of the spinner landing on each colour.
| Colour | blue | green | purple | red | yellow |
|---|---|---|---|---|---|
| Probability | 0.12 | 0.20 | 0.38 | \(4x\) | \(x\) |
Sophie spins the spinner once.
(a) Work out the probability that the spinner lands on blue or on green or on purple. (1)
Max spins the spinner 350 times.
(b) Work out an estimate for the number of times the spinner lands on red. (4)
| Scheme | Marks |
|---|---|
| 0.7 | B1 |
| (1) |
Notes
B1: oe eg \(\dfrac{7}{10}\) oe or 70% or \(\dfrac{0.7}{1}\)
If probabilities are given as percentages then % sign must be seen
| Scheme | Marks |
|---|---|
| eg 1 – (0.12 + 0.2 + 0.38) (= 0.3) oe or 1 – “0.7” (= 0.3) oe or 0.12 + 0.20 + 0.38 + 4\(x\) + \(x\) = 1 oe or “0.7” × 350 (= 245) oe or 0.12 × 350 (= 42) or 0.38 × 350 (= 133) | M1 |
| eg “0.3” ÷ 5 (= 0.06) or “0.3” ÷ 5 × 4 (= 0.24) or 0.24 or (\(x\) =) 0.06 or (4\(x\) =) 0.24 or “0.3” × 350 (= 105) oe or 350 – “245” (= 105) oe or 350 – “42” – 0.20 × 350 – “133” (=105) oe | M1 |
| eg “0.06” × 350 (= 21) oe or “105” ÷ 5 (= 21) oe or “0.06” × 4 × 350 oe or “0.24” × 350 | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 84 | A1 |
| (4) | |
| (5 marks) |
Notes
M1: ft their “0.7”
If probabilities are given as percentages then % sign must be seen
M1: or for \(\dfrac{21}{350}\) or \(\dfrac{84}{350}\)
A1: cao