Higher November 2017 Paper 5 Q14
14 Adam has 10 sweets in a bag.
5 are cherry sweets, 4 are lemon sweets and 1 is an orange sweet.
Adam chooses a sweet at random from the bag and eats it.
He then takes another sweet at random from the bag and eats it.
(a) Adam says
The probability that I choose two cherry sweets is \(\dfrac{25}{100}\).
He is incorrect. Explain his error. [2]
(b) Find the probability that the two sweets he chooses have different flavours. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| It should have been \(\dfrac{5}{10} \times \dfrac{4}{9}\) oe isw | 2 | M1 for showing \(\dfrac{5}{10} \times \dfrac{5}{10}\) or \(\dfrac{1}{2} \times \dfrac{1}{2}\) or for explaining that he did not take account that there was one less sweet for the second choice oe | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{58}{90}\) oe | 4 | M3 for \(\left(\dfrac{5}{10} \times \dfrac{5}{9}\right) + \left(\dfrac{4}{10} \times \dfrac{6}{9}\right) + \left(\dfrac{1}{10} \times \left[\dfrac{9}{9}\right]\right)\) oe | oe \(2\left(\dfrac{5}{10} \times \dfrac{4}{9}\right) + 2\left(\dfrac{4}{10} \times \dfrac{1}{9}\right) + 2\left(\dfrac{5}{10} \times \dfrac{1}{9}\right)\) or \(1 - \left(\dfrac{5}{10} \times \dfrac{4}{9}\right) - \left(\dfrac{4}{10} \times \dfrac{3}{9}\right)\) accept equivalents over 90 throughout for method and grouping of products |
| or M2 for the sum of any 2 of the above products oe isw or M1 for any correct product from above oe isw | or M2 for the sum of any 4 of the above products oe isw or M1 for any the sum of any 2 of the above products oe isw | ||
| If 0 scored, SC1 for 58 different options soi | Implied by \(\dfrac{58}{100}\) | ||