S1 January 2010 Q4
4. There are 180 students at a college following a general course in computing. Students on this course can choose to take up to three extra options.
112 take systems support,
70 take developing software,
81 take networking,
35 take developing software and systems support,
28 take networking and developing software,
40 take systems support and networking,
4 take all three extra options.
A student from the course is chosen at random.
Find the probability that this student takes
Students who want to become technicians take systems support and networking. Given that a randomly chosen student wants to become a technician,
| Scheme | Marks |
|---|---|
 | |
| 3 closed curves and 4 in centre | M1 |
| Evidence of subtraction | M1 |
| 31,36,24 | A1 |
| 41,17,11 | A1 |
| Labels on loops, 16 and box | B1 |
| (5) |
Notes
2nd M1 There may be evidence of subtraction in “outer” portions, so with 4 in the centre then 35, 40, 28 (instead of 31,36,24) along with 33, 9, 3 can score this mark but A0A0
N.B. This is a common error and their “16” becomes 28 but still scores B0 in part (a)
| Scheme | Marks |
|---|---|
| P(None of the 3 options) \(= \dfrac{16}{180} = \dfrac{4}{45}\) | B1ft |
| (1) |
Notes
B1ft for \(\tfrac{16}{180}\) or any exact equivalent. Can ft their “16” from their box. If there is no value for their “16” in the box only allow this mark if they have shown some working.
| Scheme | Marks |
|---|---|
| P(Networking only) \(= \dfrac{17}{180}\) | B1ft |
| (1) |
Notes
B1ft ft their “17”. Accept any exact equivalent
| Scheme | Marks |
|---|---|
| P(All 3 options/technician) \(= \dfrac{4}{40} = \dfrac{1}{10}\) | M1 A1 |
| (2) | |
| (9 marks) |
Notes
If a probability greater than 1 is found in part (d) score M0A0
M1 for clear sight of \(\dfrac{\mathrm{P}(S \cap D \cap N)}{\mathrm{P}(S \cap N)}\) and an attempt at one of the probabilities, ft their values.
Allow P(all 3 | \(S \cap N\)) = \(\dfrac{4}{36}\) or \(\dfrac{1}{9}\) to score M1 A0.
Allow a correct ft from their diagram to score M1A0 e.g. in 33,3,9 case in (a): \(\tfrac{4}{44}\) or \(\tfrac{1}{11}\) is M1A0
A ratio of probabilities with a product of probabilities on top is M0, even with a correct formula.
A1 for \(\dfrac{4}{40}\) or \(\dfrac{1}{10}\) or an exact equivalent
Allow \(\dfrac{4}{40}\) or \(\dfrac{1}{10}\) to score both marks if this follows from their diagram, otherwise some explanation (method) is required.