When Alice spends the day with the babysitter, there is a $0.6$ probability that she turns on the TV and watches a show. Her little sister Betty cannot turn on the TV by herself. But once the TV is on, Betty watches with probability $0.8$. Tomorrow the girls spend the day with the babysitter.
a) What is the probability that both Alice and Betty watch TV tomorrow?
b) What is the probability that Betty watches TV tomorrow?
c) What is the probability that only Alice watches TV?
Define events $A,B$ as the event that Alice turns on the TV and watches a show, and the event that Betty watches TV.
Then for a), we get $P(A\cap B)=P(A)P(B|A)=0.6\cdot 0.8=0.48.$
For b), we have $P(A|B)=1,$ so that $P(B)=\frac{P(A\cap B)}{P(A|B)}=\frac{0.48}{1}=0.48.$
For c), we need to find $P(A|B^{C})=\frac{P(A\cap B^{C})}{P(B^{C})}= \frac{1-0.48}{1-0.48}=1.$
Is any of the above correct? Thank you for time.
a) and b) are correct. Since Betty can't watch TV unless Alice does, the probabilities should be the same.
For c), you want $P(A\cap B^c)$, not $P(A\mid B^c)$. Now $P(A\cap B^c)=P(A)-P(A\cap B)=0.6-0.48=0.12$.