Wheelworks offers two repair-warranty schemes for wheel customers. Under Scheme 1, the probability a wheel needs a free repair within a year is \(0.15\). Un...
Question 1 Report
Wheelworks offers two repair-warranty schemes for wheel customers. Under Scheme 1, the probability a wheel needs a free repair within a year is \(0.15\). Under Scheme 2, the probability a wheel fails an initial inspection is \(0.4\); if it fails, the probability it then needs a free repair is \(0.5\), and if it passes, that probability is \(0.05\).
- Use the tree diagram to find the overall probability a Scheme 2 wheel needs a free repair. (3)
- A mechanic services 200 wheels under each scheme in a year. Find the expected repairs under each. (2)
- State, with a reason, which scheme is likely to cost more in free repairs. (2)
Answer Details
The overall probability of an event that can happen along two different paths of a tree diagram is the sum of the probabilities of each path, and multiplying by the number of wheels serviced then converts a probability into an expected count.
(a) Following the "fail, then repair" branch: \( P(\text{fail, repair}) = 0.4 \times 0.5 = 0.2 \) [1 mark]. Following the "pass, then repair" branch: \( P(\text{pass, repair}) = 0.6 \times 0.05 = 0.03 \) [1 mark]. Since a wheel needing repair happens via exactly one of these two paths, the overall probability is their sum: \( 0.2+0.03=0.23 \) [1 mark].
(b) Multiplying each scheme's probability of needing a repair by the 200 wheels serviced gives the expected number of repairs: Scheme 1, \( 200 \times 0.15 = 30 \); Scheme 2, \( 200 \times 0.23 = 46 \) [2 marks].
(c) Since \( 46 > 30 \), Scheme 2 is expected to need more free repairs over the year, so it is likely to cost the mechanic more [2 marks].
Even though most wheels pass Scheme 2's inspection (60% of them), the 40% that fail have a relatively high 50% repair rate, which is enough to push Scheme 2's overall repair probability above Scheme 1's flat 15% rate.
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