Here it is given atleast 1 computer is sold.
Now, we have to find the probability that A is sold
So, here 3 chances
1) A sold
2) B sold
3) both sold
So, among these 3 chances probability of A sold= $\frac{1}{3}\times \frac{40}{100}$
Probability of B sold=$\frac{1}{3}\times \frac{60}{100}$
Probability of both sold=$\frac{1}{3}\times 1$
So,total probability= $\frac{\frac{1}{3}\times \frac{40}{100}+\frac{1}{3}\times 1}{\frac{1}{3}\times \frac{40}{100}+\frac{1}{3}\times \frac{60}{100}+\frac{1}{3}\times 1}$=$\frac{7}{10}$