S1 June 2014 Q4
4. In a factory, three machines, \(J\), \(K\) and \(L\), are used to make biscuits.
Machine \(J\) makes 25% of the biscuits.
Machine \(K\) makes 45% of the biscuits.
The rest of the biscuits are made by machine \(L\).
It is known that 2% of the biscuits made by machine \(J\) are broken, 3% of the biscuits made by machine \(K\) are broken and 5% of the biscuits made by machine \(L\) are broken.
A biscuit is selected at random.
| Scheme | Marks |
|---|---|
 | M1 A1 |
| (2) |
Notes
Allow fractions or percentages throughout this question
Allow 3+6 tree diagram with the 6 correct “end” probs and labels to get 2/2 (1st, 3rd, 5th gets M1)
M1 for (3+6) tree drawn with 0.25, 0.45, 0.02, 0.03, 0.05 on correct branches
A1 for 0.3, 0.98, 0.97, 0.95 on the correct branches and labels, condone missing \(B^{\prime}\)s
| Scheme | Marks |
|---|---|
| \(0.25\times 0.98\), \(= \mathbf{0.245}\) (or exact equiv. e.g. \(\frac{49}{200}\)) | M1A1 |
| (2) |
Notes
Correct answer only scores full marks for parts (b), (c) and (d)
When using “their probability \(p\)” for M1 and A1ft they must have \(0 \lt p \lt 1\)
M1 for \(0.25\times\) ‘their 0.98’ o.e.
| Scheme | Marks |
|---|---|
| \(0.25\times 0.02 + 0.45\times 0.03 + 0.3\times 0.05\), \(= \mathbf{0.0335}\) (or exact equiv. e.g. \(\frac{67}{2000}\)) | M1A1 |
| (2) |
Notes
M1 for \(0.25\times\) their \(0.02 + 0.45\times\) their \(0.03 +\) their \(0.3\times\) their 0.05 Condone 1 transcription error.
Or \(1 - (0.25\times\) their \(0.98 + 0.45\times\) their \(0.97 +\) their \(0.3\times\) their \(0.95)\)
| Scheme | Marks |
|---|---|
| \([\mathrm{P}(J \cup L \mid B)] = \dfrac{0.25\times 0.02 + 0.3\times 0.05}{0.0335}\) or \(\dfrac{0.0335 - 0.45\times 0.03}{0.0335}\) | M1A1ft |
| \(= 0.5970\ldots\) awrt 0.597 (or \(\frac{40}{67}\) or exact equiv.) | A1 |
| (3) | |
| (9 marks) |
Notes
M1 for use of conditional probability with their (c) as denominator. Also exactly 2 products on num’ and at least one correct (or correct ft) or their (c) – one of the products from their (c). Ignore an incorrect expression inside their probability statement
A1ft for \(\dfrac{0.25\times\text{their } 0.02 + \text{their } 0.3\times\text{their } 0.05}{\text{their (c)}}\) or \(\dfrac{\text{their (c)} - 0.45\times\text{their } 0.03}{\text{their (c)}}\) or \(\dfrac{0.02}{\text{their (c)}}\)
A1 awrt 0.597 or exact fraction e.g. \(\frac{40}{67}\)