S1 June 2014 (R) Q7

EdexcelOld spec11 marksProbability

7. In a large company,
78% of employees are car owners,
30% of these car owners are also bike owners,
85% of those who are not car owners are bike owners.

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

An employee is selected at random.

(b) Find the probability that the employee is a car owner or a bike owner but not both. (2)

Another employee is selected at random.
Given that this employee is a bike owner,

Two employees are selected at random.

(d) Find the probability that only one of them is a bike owner. (3)

Scheme	Marks
	B1 B1 B1
	(3)

Mark scheme (b)

Scheme	Marks
P(car or bike but not both) \(= 0.78 \times 0.70 + 0.22 \times 0.85 = 0.733\)	M1 A1
	(2)

Notes

M1 for correct expression of follow through their correct tree branches

A1 for 0.733 or exact equivalent e.g. \(\frac{733}{1000}\) and allow 73.3%

Mark scheme (c)

Scheme	Marks
\([\mathrm{P}(\text{car} \mid \text{bike})] = \dfrac{\mathrm{P}(\text{car} \cap \text{bike})}{\mathrm{P}(\text{bike})} = \dfrac{0.78 \times 0.30}{0.78 \times 0.30 + 0.22 \times 0.85},\ = 0.555819\ldots\)	M1A1
awrt 0.556	A1
	(3)

Notes

M1 for a correct expression correct ft or correct formula and \(\dfrac{\text{1 product}}{\text{sum of 2 products}}\)
With at least 2 products correct or correct ft. Ratio must be smaller than 1

1st A1 for finding the denominator correctly. Fully correct expression or = 0.421 (oe)

2nd A1 for awrt 0.556 or exact equivalent e.g. \(\frac{234}{421}\) and allow 55.6%

Mark scheme (d)

Scheme	Marks
\(\mathrm{P}(\text{bike}) = 0.78 \times 0.30 + 0.22 \times 0.85 = 0.421\), P(not bike) = 1 – 0.421	M1
\(0.421 \times 0.579 + 0.579 \times 0.421\)	dM1
\(= 0.487518\) awrt 0.488	A1
	(3)
	(11 marks)

Notes

M1 for their \(\mathrm{P}(\text{bike}) \times (1 - \mathrm{P}(\text{bike}))\)

dM1 for \(\times 2\)

A1 for awrt 0.488