A bag contains a total of 20 books on physics and mathematics, Any possible combination of books is equally likely. Ten books are chosen from the bag and it is found that it contains 6 books of mathematics. Find out the probability that the remaining books in the bag contains 3 books on mathematics.

Explanation

Let `E_(i)(i=0,1,2,...,20)` be the event that the bag contains I books on mathematics. Since all these events are equally likely and mutually exclusive and exhaustive, so `P(E_(1))=1//21(i=0,1,2,....,20)` and let A be the event that a drw of 10 books contains 6 books on matematics Then
So, using Bay's theorem,
`P(A)=underset(i=0)overset(20)sumP(E_(i)).P(A//E_(i))`
`=1/21[underset(i=0)overset(20)sumP(A//E_(i))]`
`=1/21[underset(i=6)overset(16)sum(""^(i)C_(6)xx""^(20-i)C_(4))/(""^(20)C_(10))]`
Now, we want that the bag should contain 2 more books on mathematics, i.e., `E_(8)` must occur.
Using Baye's theorem,
`P(E_(8)//A)=(P(E_(8))P(A//E_(8)))/(P(A))`
`=((""^(8)C_(6)xx""^(12)C_(4))/(""^(20)C_(10)))/(underset(i=6)overset(16)sum((""^(i)C_(6)xx""^(20-i)C_(4))/(""^(20)C_(10))))`
`=(""^(8)C_(6)xx""^(12)C_(4))/(underset(i=6)overset(16)sum(""^(i)C_(6)xx""^(20-i)C_(4)))`