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Bayes' Theorem Problem

  1. Sep 23, 2016 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant 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]

    3. 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?
     
  2. jcsd
  3. Sep 23, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
  4. Sep 23, 2016 #3
    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.
     
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