# Probability of computer A being sold

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?

• Yes, it is correct.
– mfl
Jul 10, 2017 at 13:15
• @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? Jul 10, 2017 at 13:17
• 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 $P(A\cap B)=P(A)P(B).$
– mfl
Jul 10, 2017 at 13:19
• @mfl, thanks you very much. :) Jul 10, 2017 at 13:23