Bayes' Theorem for Employee Napping Probability

[temaire]

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:
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)}

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

P(A_1)=\frac{4}{7}
P(A_2)=\frac{3}{7}
P(B|A_1)=\frac{2}{7}
P(B|A_2)=\frac{1}{7}

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)}
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}}
P(A_1|B)=\frac{8}{11}=0.73

Is my solution correct?

 

[mfb]

Insufficient information.

You assumed that the starting time of the nap is random with a uniform distribution.
a) That should be specified in the problem.
b) That can lead to problems if the nap starts after 3 PM.

If we use that assumption of a uniform nap starting time and ignore the potential issue with 4 pm, your answer is right.

 

[temaire]

Thanks for the response.

I made the assumption that the naps could be broken up into multiple segments and that the total nap times were 2 hours or 1 hour. This avoids the problem of the employee taking an hour long nap right before his shift ends.