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Homework Help: Conditional probability question

  1. Jan 14, 2006 #1
    An insurance company runs three offices, A, B and C. The company's employess are distirbuted as follows; 30% work in office A, 20% in Off. B and 50% in Off. C.
    In office A 10% are managers, in office B 20% are managers and in office C 5% are managers

    a. What is the total proportion of managers in the company?
    b. If a member of staff, randomly chosen, turns out to be a manager, what is the probability that she works in office A.

    I've worked out part a, (0.3*0.1)+(0.2*0.2)+(0.5*0.05)=0.03+0.04+0.025=0.0 95 or 9.5%

    I'm uncertian about part b though, I think it has something to do with conditional probability which is p(b given a)=p(a and b) divided by p(a), I think.

    Thanks for the help. Edit: I did probability after calculus, but I don't know about the rest of you. That's why I posted it here.
     
    Last edited: Jan 14, 2006
  2. jcsd
  3. Jan 14, 2006 #2

    arildno

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    Well, you know the chance of a person being a manager; you also know the chance of the person being BOTH a manager and in office A, so you've got what you need to find out the chance of a manager being in A..
     
  4. Jan 14, 2006 #3
    So is this right?

    p(a manager and in office A)=0.3*0.1=0.03

    p(a manager)=0.095

    0.03/0.095=0.31579
     
  5. Jan 14, 2006 #4

    arildno

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    Should be right.
     
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