Foundation November 2024 Paper 2 Q18
18 Jack has ten cards numbered 11 to 20.
He picks a card at random.
Jack says,
In these ten cards, there are two multiples of 5 and five even numbers.
Therefore, the probability that I pick a card that is a multiple of 5 or an even number is \(\dfrac{2}{10} + \dfrac{5}{10} = \dfrac{7}{10}\).
Describe the error in Jack’s method and give the correct answer.
The error is ..................................................................
The correct answer is .............................. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| The two events are not mutually exclusive oe and \(\dfrac{6}{10}\) oe | 2 | B1 for correct reason or \(\dfrac{6}{10}\) oe | e.g. He has counted the same card/number/20 twice It should be \(\dfrac{1}{10} + \dfrac{5}{10}\) or \(\dfrac{2}{10} + \dfrac{4}{10}\) No contradictory statements for the reason See appendix 2 |
Appendix
APPENDIX 2 Question 18
| Response | Mark | |
|---|---|---|
| A | Some multiples of 5 may intersect with the five even numbers so it may not be \(\dfrac{5}{10}\) BOD | 1 |
| B | One of the multiples of 5 could be even BOD | 1 |
| C | He has counted the same card more than once | 1 |
| D | 20 is repeated/in both | 1 |
| E | He did not minus the repeated card | 1 |
| F | He forgot to take away the multiples of 5 from the even numbers | 1 |
| G | They have counted 20 twice and they have not added the fractions correctly (we can ignore the second part of the statement as it does not contradict they have counted 20 twice) | 1 |
| H | Some even numbers are also numbers that are multiples of 5 | 1 |
| I | Even numbers are also numbers that are multiples of 5 (not true – only some even numbers are) | 0 |
| J | The two multiples of 5 are also even numbers (incorrect statement) | 0 |