Question

The probability on any given night that it's Abe’s responsibility to cook dinner is 24%. If it’s Abe’s responsibility to cook dinner, the probability that his family goes out to a restaurant to eat is 65%. If it is not Abe’s responsibility to cook dinner, the family goes to a restaurant only 15% of the time. Create a tree diagram for this situation: What is the probability that Abe’s family eats out on a night that Abe was not responsible to cook dinner?On any given night, what is the probability Abe’s family eats at a restaurant?Abe’s family did not eat at a restaurant. Determine the probability that Abe was not responsible for cooking?

Answer 1

Let A be the event that "it's Abe's responsibility to cook dinneron any given night" and B be the event that "family goes out to a restauarnt to eat".

iven that:

`\begin{gathered} P(A)=0.24 \\ P(B|A)=0.65 \\ P(B|A^C)=0.15 \end{gathered}`

Draw the tree diagram.

Use Baye's theorem

`P(A|B)=(P(A\cap B))/(P(B))`

`\begin{gathered} \text{P(Abe's family eats out on a night and Abe was not responsible to cook) } \\ =P(B\cap A^C) \\ =P(B|A^C)\cdot P(A^C) \\ =0.15(0.76) \\ =0.114 \end{gathered}`