Higher November 2019 Unit 1 Q18
18. Sixteen balls are placed in a bag.
Ten of the balls are green and six are yellow.
Two balls are selected at random and not replaced.
Benjamin states that the probability of selecting two balls of the same colour is equal to the probability of selecting two balls of different colours.
Is Benjamin correct?
You must show all your working to justify your answer. [4]
| Scheme | Marks | Comments |
|---|---|---|
| (P[same colour] =) 10/16 × 9/15 + 6/16 × 5/15 or equivalent OR (P[different colours] =) 10/16 × 6/15 + 6/16 × 10/15 or equivalent. | M2 | M1 for sight of any correct product. |
| = 120/240 or equivalent | A1 | Award for the answer to either probability (total). Mark final answer. Do not ignore incorrect cancelling. If both probabilities are evaluated, accept 240 written as 16 × 15. If M0A0, award SC1 for an answer of 136/256 or 120/256 (method ‘with replacement’). |
| ‘Yes’ with explanation (must refer to the ‘other’ probability) | E1 | If M2A0 or SC1 awarded, then award E1 for ‘No’, provided only one answer evaluated (from calculating products), and a valid explanation given based on P[same colour] + P[different colours] = 1 or E1 for ‘Yes’ if both probabilities (incorrectly) evaluated and ‘their P[same colour]’ = ‘their P[different colours]’ E0 if both probabilities evaluated and ‘their P[same colour]’ + ‘their P[different colours]’ ≠ 1 with ‘their P[same colour]’ ≠ ‘their P[different colours]’. |
Corrected by Westie's Workshop
- The board's mark scheme printed “(method ‘without replacement’)”. Changed to “(method ‘with replacement’)” because 136/256 and 120/256 come from putting the first ball back before taking the second (with replacement); the question says the balls are not replaced, which is why these answers only get a special case mark.
Original: see the board's mark scheme PDF (link above).