S1 June 2012 Q7
7. A manufacturer carried out a survey of the defects in their soft toys. It is found that the probability of a toy having poor stitching is 0.03 and that a toy with poor stitching has a probability of 0.7 of splitting open. A toy without poor stitching has a probability of 0.02 of splitting open.
The manufacturer also finds that soft toys can become faded with probability 0.05 and that this defect is independent of poor stitching or splitting open. A soft toy is chosen at random.
| Scheme | Marks |
|---|---|
 | Shape B1 Labels & 0.03 B1 Labels & 0.7, 0.02 B1 |
| (3) |
Notes
Allow MR of 0.2 for 0.02 or 0.3 for 0.03 on tree diagram to score all M and A1ft marks only
1st B1 for 2 branch then 4 branch shape
2nd dB1 dep. on 1st B1 for labels showing stitching (accept letters) and 0.03 value correctly placed
3rd dB1 dep. on 1st B1 for labels showing splitting and 0.7 and 0.02 correctly placed
[probabilities shown in brackets are not required and any such values given can be ignored in (a)]
| Scheme | Marks |
|---|---|
| P(Exactly one defect) \(= 0.03\times 0.3 + 0.97\times 0.02\) or \(\mathrm{P}(PS \cup Split) - 2\mathrm{P}(PS \cap Split)\) | M1A1ft |
| \(= [0.009 + 0.0194 =]\) 0.0284 | A1 cao |
| (3) |
Notes
M1 for \(0.03\times p + 0.02\times q\) where \(p\) and \(q\) follow from their tree diagram. Extra terms is M0
1st A1ft for a fully correct expression. Accept \(1 - 0.7\) for 0.3 and \(1 - 0.03\) for 0.97. Follow through 0.2 and 0.3 MR only
MR 0.2 for 0.02 \(\to\) 0.203 or 0.3 for 0.03 \(\to\) 0.104 or both \(\to\) 0.23 should score M1A1A0
2nd A1 cao for 0.0284 only (or exact equivalent such as \(\frac{71}{2500}\))
| Scheme | Marks |
|---|---|
| P(No defects) \(= (1 - 0.03)\times(1 - 0.02)\times(1 - 0.05)\) (or better) | M1 |
| \(= 0.90307\) awrt 0.903 | A1 cao |
| (2) |
Notes
Do not allow 0.5 as MR of 0.05 so no M or A marks in (c) or (d)
M1 for (their 0.97)\(\times\)(their 0.98)\(\times(1 - 0.05)\) (or better) f.t. values from their tree diagram
A1 cao for awrt 0.903
| Scheme | Marks |
|---|---|
| P(Exactly one defect) \(= \text{(b)}\times(1 - 0.05) + (1 - 0.03)\times(1 - 0.02)\times 0.05\) | M1 M1 |
| \(= \text{"}0.0284\text{"}\times 0.95 + 0.97\times 0.98\times 0.05\) | A1ft |
| \(= [0.02698 + 0.04753] = 0.07451\) awrt 0.0745 | A1 cao |
| (4) | |
| (12 marks) |
Notes
1st M1 for one correct triple (or correct ft from their tree) of:
\([0.03\times 0.3\times(1 - 0.05)] + [0.97\times 0.02\times(1 - 0.05)] + [0.97\times 0.98\times 0.05]\)
2nd M1 for two correct triples or correct ft from their tree and adding or their (b) \(\times(1 - 0.05)\)
1st A1ft for a fully correct expression or f.t. their (b) and 0.2 or 0.3 MR only
MR 0.2 for 0.02 \(\to\) 0.23165 or 0.3 for 0.03 \(\to\) 0.1331 or both \(\to\) 0.2465 (or awrt 3sf) scores M1M1A1A0
2nd A1 cao for awrt 0.0745