Foundation November 2019 Paper 3 Q16
16 In a bag there are only red counters, blue counters, green counters and yellow counters.
A counter is taken at random from the bag.
The table shows the probabilities of getting a red counter or a yellow counter.
| Colour | red | blue | green | yellow |
|---|---|---|---|---|
| Probability | 0.4 | \(\ldots\ldots\) | \(\ldots\ldots\) | 0.25 |
the number of blue counters : the number of green counters = 3 : 4
Complete the table. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| blue 0.15 green 0.2 | P1 | for \(1 - 0.4 - 0.25\ (=0.35)\) oe |
| P1 | for using the ratio, eg \(\text{``}0.35\text{''} \div (3 + 4)\ (=0.05)\) or \(\text{``}0.35\text{''} \times \frac{3}{7}\ (=0.15)\) or \(\text{``}0.35\text{''} \times \frac{4}{7}\ (=0.2)\) | |
| P1 | for a complete process \(3 \times \text{``}0.05\text{''}\ (=0.15)\) and \(4 \times \text{``}0.05\text{''}\ (=0.2)\) or \(\text{``}0.35\text{''} - \text{``}0.15\text{''}\ (=0.2)\) or \(\text{``}0.35\text{''} - \text{``}0.2\text{''}\ (=0.15)\) or green 0.15, blue 0.2 | |
| A1 | oe |
Additional guidance
May work in percentages, condone missing % sign
If the two numbers in the table sum to 0.35 that implies P1
One correct value in the table implies P2
7 can come from 3+4
Accept answers given in decimals, fractions or percentages.