Higher June 2022 Paper 1 Q2
2 Here is a biased 4-sided spinner.
The table gives the probabilities that, when the spinner is spun once, it will land on 1 or it will land on 3
| Number | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Probability | 0.26 | 0.18 |
The probability that the spinner will land on 2 is equal to the probability that the spinner will land on 4
Ravina is going to spin the spinner a number of times.
Ravina works out that an estimate for the number of times the spinner will land on 3 is 45
Work out an estimate for the number of times the spinner will land on 4
(4)
| Scheme | Marks |
|---|---|
| 1 – (0.26 + 0.18) (= 0.56) oe or 0.28 oe or \(x + x = 1 - (0.26 + 0.18)\) oe | M1 |
45 ÷ 0.18 (= 250) oe or \(\dfrac{45}{18}\) (= 2.5) oe \(\dfrac{\text{``}{0.56}\text{''}}{2} \div 0.18\left(= \dfrac{14}{9} = 1.55\ldots\right)\) oe or \(\dfrac{\text{``}{56}\text{''}}{2} \div 18\left(= \dfrac{14}{9} = 1.55\ldots\right)\) | M1 |
\(\text{``}{250}\text{''} \times \dfrac{\text{``}{0.56}\text{''}}{2}\) oe or \(2.5 \times \dfrac{\text{``}{56}\text{''}}{2}\) oe or \(\text{``}{250}\text{''} \times \text{``}{0.28}\text{''}\) oe or \(\text{``}{0.28}\text{''} \div 0.18 \times 45\) oe or \(\text{``}{\dfrac{14}{9}}\text{''} \times 45\) oe or “28” ÷ 18 × 45 oe or \(\dfrac{45}{18} \times \text{``}{28}\text{''}\) oe | M1 |
| 70 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: 0.28 oe may be seen in the table
A1: (\(\dfrac{70}{250}\) scores M3A0)