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Date May 2011 Marks available 4 Reference code 11M.2.sl.TZ1.7
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 7 Adapted from N/A

Question

A company uses two machines, A and B, to make boxes. Machine A makes \(60\% \) of the boxes.

\(80\% \) of the boxes made by machine A pass inspection.

\(90\% \) of the boxes made by machine B pass inspection.

A box is selected at random.

Find the probability that it passes inspection.

[3]
a.

The company would like the probability that a box passes inspection to be 0.87.

Find the percentage of boxes that should be made by machine B to achieve this.

[4]
b.

Markscheme

evidence of valid approach involving A and B     (M1)

e.g. \({\rm{P}}(A \cap {\rm{pass}}) + {\rm{P}}(B \cap {\rm{pass}})\) , tree diagram

correct expression     (A1)

e.g. \({\rm{P}}({\rm{pass}}) = 0.6 \times 0.8 + 0.4 \times 0.9\)

\({\rm{P}}({\rm{pass}}) = 0.84\)     A1     N2

[3 marks]

a.

evidence of recognizing complement (seen anywhere)     (M1)

e.g. \({\rm{P}}(B) = x\) , \({\rm{P}}(A) = 1 - x\) , \(1 - {\rm{P}}(B)\) , \(100 - x\) , \(x + y = 1\)

evidence of valid approach     (M1)

e.g. \(0.8(1 - x) + 0.9x\) , \(0.8x + 0.9y\)

correct expression     A1

e.g. \(0.87 = 0.8(1 - x) + 0.9x\) , \(0.8 \times 0.3 + 0.9 \times 0.7 = 0.87\) , \(0.8x + 0.9y = 0.87\)

\(70\% \) from B     A1     N2

[4 marks]

b.

Examiners report

Part (a) was usually well done. Those candidates that did not succeed with this part often did not show a correct tree diagram indicating that they did not really understand the problem or indeed how to start it.

a.

Many successful attempts to (b) relied on "guess and check" or intuitive solutions while a surprising number of candidates could not manage to systematically set up an appropriate algebraic expression involving a complement.

b.

Syllabus sections

Topic 5 - Statistics and probability » 5.6 » Combined events, \(P\left( {A \cup B} \right)\) .
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