Conditional probability question

  • Thread starter fobster
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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.
 
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  • #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..
 
  • #3
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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
 
  • #4
arildno
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Should be right.
 

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