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There are 2 white and 4 black balls in urn A; in urn B, there are 4 white and 7 black balls. If one ball is randomly replaced from A into B and a ball is drawn from B then find the probability for the ball to be a white one?
13/36=0.36?
is it 17/36?
$\frac{2}{6}\times \frac{5}{12}+\frac{4}{6}\times \frac{4}{12}=\frac{13}{36}$

There are possible scenarios :

A) White ball is replaced from A into B  and  white ball is picked up :   (1 / 3) * (5 / 12)  =  5 / 36

B) White ball is replaced from A into B  and  black ball is picked up :   (1 / 3) * (7 / 12)  =  7 / 36

C) Black ball is replaced from A into B and white ball is picked up :      (2 / 3) * (4 / 12)  =  8 / 36

D) Black ball is replaced from A into B and black ball is picked up :      (2 / 3) * (8 / 12)  =  16 / 36

So P(ball picked will be white one) :        = [ A) + C) ] / [ A) + B) + C) + D) ]

= (5 + 8) / (5 + 7 + 8 + 16)

= 13 / 36

Hence the correct answer is 13 / 36