Higher June 2025 Unit 1 Q20
20. Mathew has six cards.
He chooses three of the cards at random.
He does not replace any of the cards.
(a) Calculate the probability that the first card is a 7, the second card is a 5 and the third card is a 6. [2]
(b) Calculate the probability that the product of the numbers on Mathew’s three cards is a cube number. [2]
(c) Calculate the probability that the sum of the numbers on Mathew’s three cards is 17. [3]
| Scheme | Marks | Comments |
|---|---|---|
| [P(7,5,6) =] \(\dfrac{1}{6}\) × \(\dfrac{4}{5}\) × \(\dfrac{1}{4}\) or equivalent | M1 | |
| = \(\dfrac{4}{120}\) (= \(\dfrac{1}{30}\)) or equivalent | A1 | ISW (for incorrect cancelling or conversion). No marks for working with replacement. |
| Scheme | Marks | Comments |
|---|---|---|
| [P(5,5,5) =] \(\dfrac{4}{6}\) × \(\dfrac{3}{5}\) × \(\dfrac{2}{4}\) or equivalent | M1 | FT their denominator from (a), provided it is derived from the product of 3 consecutive integers or it is 216. |
| = \(\dfrac{24}{120}\) (= \(\dfrac{1}{5}\)) or equivalent | A1 | ISW (for incorrect cancelling or conversion). If no marks, award SC1 for an answer of (\(\dfrac{64}{216}\) + \(\dfrac{1}{216}\) + \(\dfrac{1}{216}\) =) \(\dfrac{66}{216}\) (= \(\dfrac{11}{36}\)) from working with replacement. |
| Scheme | Marks | Comments |
|---|---|---|
| [P(5,5,7 or 5,7,5 or 7,5,5) or 3 × P(5,5,7) =] \(\dfrac{4}{6}\) × \(\dfrac{3}{5}\) × \(\dfrac{1}{4}\) × 3 or equivalent | M2 | FT their denominator from (a), provided it is derived from the product of 3 consecutive integers or it is 216. Award M1 for:
|
| = \(\dfrac{36}{120}\) (= \(\dfrac{3}{10}\)) or equivalent | A1 | CAO. ISW (for incorrect cancelling or conversion). If no marks, and if working with replacement: award SC2 for one of the following answers
award SC1 for one of the following answers
|