Qn:Two balls are selected sequentially from an urn containing six red, three white, and four blue balls. What is the probability of selecting a white ball on the second draw if the first ball is not replaced before the second is selected?
The first ball can be white or non-white(red or blue).
case 1:The first ball is white: Then the probability of second ball is white is $3/13*2/12 = 1/26$.
case 2:The first ball is non-white: Then the probability of second ball is white is $10/13*3/12 = 5/26$.
So the total probability is $ 1/26 + 5/26 = 3/13$. Is this correct?