Higher June 2025 Paper 5 Q21
21 There are 10 sweets in a bag.
4 of the sweets are red, 4 are green and 2 are yellow.
(a) Charlie says,
The probability of picking the two yellow sweets at random from the bag can be found by working out \(\frac{2}{10} \times \frac{2}{10}\).
Write down the assumption that Charlie has made. [1]
(b) Mia takes one of the 10 sweets from the bag at random and eats it.
Ben then takes a sweet from the bag and eats it.
Ben then takes a sweet from the bag and eats it.
Find the probability that the bag now holds more red sweets than green sweets.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| The two events are independent or probability of yellow stays the same or there are 10 sweets in the bag for the 2nd pick OR Puts the first sweet back oe e.g. the sweets are replaced | 1 | Do not accept incorrect statements e.g. You do not pick a yellow sweet first and if you do you put it back in the bag | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{28}{90}\) oe with correct working | 5 | Correct working requires evidence of at least M4 isw cancelling/conversion to dec/% after \(\frac{28}{90}\) leading to answer For method marks throughout e.g. RY = \(\frac{4}{10} \times \frac{2}{9}\) ignore label RY | |
| M4 for \(\frac{4}{10} \times \frac{2}{9} + \frac{2}{10} \times \frac{4}{9} + \frac{4}{10} \times \frac{3}{9}\) oe | M4 accept \(\frac{8}{90} + \frac{8}{90} + \frac{12}{90}\) or \(\frac{4}{45} + \frac{4}{45} + \frac{6}{45}\) oe | ||
| or M3 for the addition of two of the above products oe (no extras) or all three products oe shown with no more than one extra product oe | M3 accept e.g. \(\frac{8}{90} + \frac{12}{90}\) or \(\frac{4}{45} + \frac{6}{45}\) oe | ||
| or M2 for one or more of the above products (standalone or added to extras) | M2 accept e.g. \(\frac{8}{90}\) or \(\frac{4}{45}\) or \(\frac{12}{90}\) or \(\frac{6}{45}\) oe nfww M3 and M2 spoiled if part of a larger product with other probabilities | ||
| or M1 for only GY, YG and GG identified or for \(\frac{4}{10}\) and \(\frac{n}{9}\) seen or for \(\frac{2}{10}\) and \(\frac{n}{9}\) seen | M1 may be seen in a list or indicated on a tree Where \(0 \lt n \lt 9\) | ||
| or for \(\frac{k}{10} \times \frac{p}{9}\) oe seen | Where \(0 \lt k \lt 10\) and \(0 \lt p \lt 9\) M1 implied by \(\frac{m}{90}\) seen , \(1 \lt m \lt 90\) | ||
| If 0 or 1 scored, instead award SC2 for answer \(\frac{28}{90}\) oe with no or insufficient working | |||