Question :- From an urn containing $a$ white balls and $b$ black balls, a certain number of balls, $k$ is drawn and they are laid aside their colour unnoticed. Then one more ball is drawn. Find the probability that it is white.
My approach:-
Let $i$ be the number of white balls drawn
Therefore, the number of black balls drawn is $=$ $k-i$
Let E be when the ball is white
$F(i)$ $=$ "$i$ white balls are drawn"
$P(E) = \sum_{i=0}^kP(E\cap F_i)$
$P(E) = P(E \cap F_0)+P(E \cap F_1)+P(E \cap F_2)+...+P(E \cap F_k)$
$P(E) = \frac{a-0}{a+b-k}+\frac{a-1}{a+b-k}+\frac{a-2}{a+b-k}+...+\frac{a-k}{a+b-k}$
$P(E) = \frac{a+a-1+a-2+a-3+...+a-k}{a+b-k} $
$P(E) = \frac{a(k+1)-(1+2+3+4+...+k)}{a+b-k} $
$P(E) = \frac{a(k+1)-\frac{k(k+1)}{2}}{a+b-k} $
$P(E) = \frac{2a(k+1)-k(k+1)}{2(a+b-k)} = \frac{(k + 1)(2a - k)}{2(a + b - k)}$
The correct answer according to the book is $\frac{a}{a+b}$
What is the best way to work through this problem?