Answer:
0.1163
Explanation:
given that if a television service subscriber has cable service, there is a 0.24 probability the subscriber has a DVR player. If the subscriber does not have cable service (e.g., has satellite service), there is a 0.7 probability the subscriber has a DVR player
Let A1 be the event subscriber has cable service and A2 subscriber does not have cable service
A1 and A2 are mutually exclusive and exhaustive
P(A1) = 0.75, P(A2) = 0.25
B = Subscriber has a DVD player
P(B/A1) =0.24 and P(B/A2) = 0.7
Prob for the subscriber does not have a DVR player
= P(B') = P(A1B')+P(A2B')
![= 0.75*(1-0.24)+0.25(1-0.7)\\= 0.57+0.075=0.645](https://img.qammunity.org/2021/formulas/mathematics/college/indofizf0rs0o1s3cdayngugzz5zqmptdb.png)
required probability = P(A2/B') = P(A2B')/P(B)
=
![(0.075)/(0.645) \\=0.1163](https://img.qammunity.org/2021/formulas/mathematics/college/swz2dze1h3bxlet7z12k5ojcq7ldurgyhe.png)