S1 January 2007 Q2

EdexcelOld spec11 marksProbability

2. In a factory, machines \(A\), \(B\) and \(C\) are all producing metal rods of the same length. Machine \(A\) produces 35% of the rods, machine \(B\) produces 25% and the rest are produced by machine \(C\). Of their production of rods, machines \(A\), \(B\) and \(C\) produce 3%, 6% and 5% defective rods respectively.

(a) Draw a tree diagram to represent this information. (3)

(b) Find the probability that a randomly selected rod is

(i) produced by machine \(A\) and is defective,

(ii) is defective. (5)

Tree diagram: A 0.35, B 0.25, C; from A, D 0.03 (0.0105); from B, D 0.06 (0.015); from C, D 0.05 (0.02); each with a D-bar branch — Mark scheme (a)

Scheme	Marks
Correct tree shape	M1
\(A\), \(B\) and \(C\) and 0.35 and 0.25	A1
\(D\) (×3) and 0.03, 0.06, 0.05 (May be implied by seeing \(\mathrm{P}(A \cap D)\) etc at the ends)	A1
	(3)

Mark scheme (b)

Scheme	Marks
(i) \(\mathrm{P}(A \cap D) = 0.35 \times 0.03, \ = \underline{\mathbf{0.0105}}\) or \(\dfrac{21}{2000}\)	M1, A1
\(\mathrm{P}(C) = 0.4\) (anywhere)	B1
(ii) \(\mathrm{P}(D) =\) (i) \(+ 0.25 \times 0.06 + (0.4 \times 0.05)\)	M1
\(= \underline{\mathbf{0.0455}}\) or \(\dfrac{91}{2000}\)	A1
	(5)

Notes

1^st M1 for 0.35×0.03. Allow for equivalent from their tree diagram.

B1 for \(\mathrm{P}(C) = 0.4\), can be in correct place on tree diagram or implied by 0.4×0.05 in \(\mathrm{P}(D)\).

2^nd M1 for all 3 cases attempted and some correct probabilities seen, including +. Can ft their tree.
Condone poor use of notation if correct calculations seen. E.g. \(\mathrm{P}(C \mid D)\) for \(\mathrm{P}(C \cap D)\).

Mark scheme (c)

Scheme	Marks
\(\mathrm{P}(C \mid D) = \dfrac{\mathrm{P}(C \cap D)}{\mathrm{P}(D)}, = \dfrac{0.4 \times 0.05}{\text{(ii)}}\)	M1, A1ft
\(= 0.43956\ldots\) or \(\dfrac{40}{91}\)	A1
	(3)
	(11 marks)

Notes

A1 0.44 or awrt 0.440

[Correct answers only score full marks in each part]

M1 for attempting correct ratio of probabilities. There must be an attempt to substitute some values in a correct formula. If no correct formula and ration not correct ft score M0.
Writing \(\mathrm{P}(D \mid C)\) and attempting to find this is M0.
Writing \(\mathrm{P}(D \mid C)\) but calculating correct ratio – ignore notation and mark ratios.

A1ft must have their 0.4×0.05 divided by their (ii).
If ratio is incorrect ft (0/3) unless correct formula seen and part of ratio is correct then M1.