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.(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Conditional probability question

**Physics Forums | Science Articles, Homework Help, Discussion**