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Homework Statement
An employee goes to work from 9 am to 4 pm. He takes a nap for an average of 2 hours if he starts napping before 1 pm and naps for an average of 1 hours if he starts napping after 1 pm. His boss randomly checks up on him once during his shift. If his boss finds him napping, what is the probability that he starts napping before 1 pm?
Homework Equations
Bayes' Theorem:
[tex]P(A_1|B)=\frac{P(A_1)P(B|A_1)}{P(B|A_1)P(A_1)+P(B|A_2)P(A_2)}[/tex]
The Attempt at a Solution
Event A1: Employee naps before 1 pm
Event A2: Employee naps after 1 pm
Event B: Boss finds employee napping
[tex]P(A_1)=\frac{4}{7}[/tex]
[tex]P(A_2)=\frac{3}{7}[/tex]
[tex]P(B|A_1)=\frac{2}{7}[/tex]
[tex]P(B|A_2)=\frac{1}{7}[/tex]
[tex]P(A_1|B)=\frac{P(A_1)P(B|A_1)}{P(B|A_1)P(A_1)+P(B|A_2)P(A_2)}[/tex]
[tex]P(A_1|B)=\frac{\frac{4}{7}\cdot \frac{2}{7}}{\frac{2}{7}\cdot \frac{4}{7}+\frac{1}{7}\cdot \frac{3}{7}}[/tex]
[tex]P(A_1|B)=\frac{8}{11}=0.73[/tex]
Is my solution correct?