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 the tv on 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 tomorrow? hint: de ne events precisely and use the product rule and the law of total probability.

A. P("Both Alice and Betty watch TV") = 12/25

B. P("Betty watches TV") = 12/25

C. P("Only Alice watches TV") = 3/25

Step-by-step explanation:

A. Because what we are told, Betty need that Alice turn on the TV, so, we first need the probability that she watches TV:

P(Alice)=3/5 (0.6)

And we know that:

P(Betty)=4/5 (0.8)

If we want the probability of both things happening at the same time (If you use a tree diagram, those events will be in the same branch), we proceed multiply them:

P("Both Alice and Betty watch TV") = 3/5 * 4/5 = 12/25

And this is the answer

B. Considering that Betty needs Alice to turn on the TV, the probability of Betty watching TV is the same as if she is with Alice.

P("Betty watches TV") = P("Both Alice and Betty watch TV") = 12/25

C. We use the same process as part A, but with a little difference. We now multiply for the probability that Betty does not watch TV (Because they still be in the same branch).

P("Betty does not watch TV") = 1 - P("Betty")

P("Betty does not watch TV") = 1 - 4/5

P("Betty does not watch TV") = 1/5

And the answer for part C is:

P("Alice watches TV without Betty") = 3/5 * 1/5 = 3/25

idk

step-by-step explanation:

the answer is 11.2 (i know i’m late)