An appliance wholesaler distributes washing machines made in factories in Akron and Wheeling. Some of these washing machines receive, at customer request, an in-home warranted repair, with the exact breakdown being; Warranted repair No Warranted repair Akron 20 160 Wheeling 50 150 If one the washing machine customers is randomly selected to check on consumer satisfaction, and A is the event that the machine was made in Akron, and R is the event that it had an in-home warranted repair, determine each of the following probabilities: a- P(A) b- P(Rc) c- P(A∩R) d- P(A|R) e- P(AcUR) f- P(Rc|A)
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1 Answer

The number of Akron machine customers is 160+20=180.

The number of Wheeling machine customers is 50+150=200.

Total number of customers is 180+200=380.

The probability of a random customer being an Akron customer is 180/380=9/19. So the probability of being a Wheeling customer is 1-9/19=10/19 (=200/380). So (a) P(A)=9/19 or 0.4737 or 47.37%.

The meaning of Rc hasn't been defined so I take it to be the complement of R, in other words, the set of customers not having the in-home repair warranty. Similarly, Ac represents the Wheeling customers.

Out of 380 customers, (20+50)=70 took up the offer of the in-home repair, so that's (b) P(R)=70/380, 0.1842 or 18.42%. The probability P(Rc) of no in-home repair warranty is 1-70/380=310/380=0.8158, 81.58%.

(c) The probability of A intersection R, that is, being an Akron customer and having the repair warranty is 20/380=1/19=0.0526, 5.26%.

(d) To find the probability of being an Akron customer or having the warranty, we need to count the customers matching these requirements. 180 Akron customers + 50 Wheeling customers having the warranty=230, so the probability is 230/380=23/38=0.6053, 60.53%.

(e) The total of Wheeling customers, some of whom have the warranty, and Akron customers having the warranty is 200+20=220. This gives the probability 220/380=11/19=0.5789, 57.89%.

(f) 180 Akron customer Service + 150 Wheeling customers with no warranty = 330, making the selection probability 330/380=0.8684, 86.84%.


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