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I have this problem:

A Ph.D. graduate has applied for a job with two 
universities: A and B. The graduate feels that she has a 
60% chance of receiving an offer from university A and a 
50% chance of receiving an offer from university B. If 
she receives an offer from university B, she believes that 
she has an 80% chance of receiving an offer from 
university A.
a) What is the probability that both universities will make 
her an offer?
b) What is the probability that at least one university will 
make her an offer?
c) If she receives an offer from university B, what is the 
probability that she will not receive an offer from 
university A?

And I have done it like this:

    A         A'
B  0.4        0.1    p(B)=0.5    

B' 0.2        0.3   p(B')=0.5
  p(A)=0.6   p(A')=0.4

p(A|B)=0.8
p(A and B)=p(B)p(A|B)=0.5(0.4)=0.4

a) p(both)=p(A and B)=0.4
b) p(at least 1)=1-p(none)= 1-p(A' and B')=0.7
c)p(A'|B)=p(B|A')p(A')/p(B)=(0.25(0.4))/0.5=0.2

p(B|A')=p(A' and B)/p(A')= 0.1/0.4=0.25

So is this the correct way of approaching it? Is it correct? Thanks.

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  • $\begingroup$ This sounds oddly familiar to your recent question here. Could you elaborate on how they differ? $\endgroup$ – Christoph_J Nov 22 '13 at 20:22
  • $\begingroup$ Yes in the previous one I had the joint of p(H and D), in this I think I have the conditional p(A|B) but I'm not sure. :) $\endgroup$ – Pedro.Alonso Nov 22 '13 at 20:46
  • $\begingroup$ It's not like I have the answers to crosscheck them with what I get, so that's why I put it here. You are the teachers :) $\endgroup$ – Pedro.Alonso Nov 23 '13 at 22:23
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So the answer is yes this is correct, there is another way to get b) without the 1-p formula like p(A or B)= p(A)+ p(B) - p(A and B), but I wanted to see the conditionals and use it. Thanks.

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