given that no of white balls = m
no of black balls = n
total no of balls = m+n
probability of 2nd ball is white:
1st case: (w,w)
P=
m
m+n
⋅
m-1
m+n-1
eqn(1)
case 2 : (b, w)
n
m+n
⋅
m
m+n-1
eqn (2)
total probability is (w,w) + (b,w):
m⋅(m-1)
(m+n)⋅(m+n-1)
+
n⋅m
(m+n)⋅(m+n-1)
=
m
(m+n)⋅m+n-1
⋅m+n-1
=
m
m+n
