rest of the times Points) al car repair garage cames out the work on time for on

Question

rest of the times Points) al car repair garage cames out the work on time for only 80% of there arJobs, and is late the rest of the time. Historical data shows that: If the repair is made on time, there is a probability of 0.75 that the repair work is satisfactory. If the repair work is done late, the probability of the work being satisfactory is 0.6. For a randomly selected repair job at the garage, find the probability of the following: (a) That the repair work is both done on-time, and is satisfactory. (b) That the repair work is neither on-time nor satisfactory

Explanation / Answer

Let Event W: Car repair garrage carries out the work on time
P(W) = 0.80
P(Wc) = 0.20
Let Event S: Repair work is satisfactory
Given: P(S | W) = 0.75
P(S | Wc) = 0.60
(a) Probability that repair work is both satisfactory and done on time:
P(W S)
P(S | W) = P(SW) / P(W)
P(SW) = P(W).P(S|W) = 0.80 x 0.75 = 0.60
(b) Probability that the repair work is neither on time nor satisfactory
P(WcSc) = 1 - P(WUS)
P(W U S) = P(W) + P(S) - P(WS)
Using total law: P(S) = P(W)P(S|W) + P(Wc)P(S|Wc)
P(S) = (0.80 x 0.75) + (0.20 x 0.60) = 0.72
P(WUS) = 0.80 + 0.72 - 0.60 = 0.92
P(WcSc) = 1 - 0.92 = 0.08

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