$\bf A(t)=942e^{-0.012t}\qquad \boxed{\stackrel{\textit{after 71 years}}{t=71}}\qquad \qquad A(t)=942e^{-0.012(71)}
\\\\\\
A(t)\approx 401.82042087707606813879$

4 0

3 years ago

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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 littl

lara [203]

Answer:

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

5 0

4 years ago