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I stumbled upon a question...

Two computers $A$ and $B$ are to be sold. A salesman who is assigned the job of selling these has the chances of $60$ percent and $40$ percent respectively to get success. The two computers may be sold independently. Given that at least one computer has been sold, the probability that computer $A$ has been sold?

Now I think of approaching this as,

$$P(A\cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.4 -0.24 = 0.76$$

$$P(A | A\cup B) = \frac {P(A \cap(A\cup B))}{P(A\cup B)} = \frac{P(A)}{P(A\cup B)} = \frac{0.6}{0.76} = 0.789$$

Is this correct?

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    $\begingroup$ Yes, it is correct. $\endgroup$
    – mfl
    Jul 10, 2017 at 13:15
  • $\begingroup$ @mfl thanks :). can you tell me one more thing? What should be P(A and B both sold)? Isn't it P(A intersection B) = .24 and this is because they are independent? $\endgroup$
    – lu5er
    Jul 10, 2017 at 13:17
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    $\begingroup$ Yes, the probability that both computers are sold is $P(A\cap B)=.24.$ As you say you can get this using the hypothesis that both events are independent. Thus $P(A\cap B)=P(A)P(B).$ $\endgroup$
    – mfl
    Jul 10, 2017 at 13:19
  • $\begingroup$ @mfl, thanks you very much. :) $\endgroup$
    – lu5er
    Jul 10, 2017 at 13:23

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