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%.