I was reading probability from Sheldon Ross and found below example.
A purchaser of electrical components buys them in lots of size 10. It is his policy to inspect 3 components from a lot and to accept the lot if all the 3 are nondefective. If 30 percent of the lots have 4 defective components and 70 percent have only 1, what proportion of lots does the purchaser reject?
And the solution is presented like this
Let A denote the event that the purchaser accepts a lot. Now,
P(A)= $P(A|lot\,has\,4\,defectives\,)\frac{3}{10}+P(A|lot\,has\,1\,defective)\frac{7}{10}$
I understood the conditional probability part but why they have multiplied it with $\frac{3}{10}$ in the first part and $\frac{7}{10}$ in the second?
