A box contains 2 red, 3 black and 4 blue balls. 3 balls are randomly drawn from the box. If two balls are found to be different, what is the probability that the third ball is blue? Select the correct option.
A) 8/3  B) 9/3  C) 5/3  D) 2/3  E) 1/3
there are 6 ways to draw a blue ball at third attempt if first two balls are different. they are   1)RED BLACK BLUE   2)RED BLUE BLUE   3)BLACK RED BLUE   4)BLACK BLUE BLUE   5)BLUE RED BLUE   6)BLUE BLACK BLUE   Probability of 1st way is (2/9)*(3/8)*(4/7)   as at first attempt we got red which are two in number out of 9 and in second we got black which are 3 in number out of 8 (as first ball is removed)and in 3rd attempt we got blue which are 4 in number out of 7 balls left.   similarly probability of other ways is as follows   (2/9)*(4/8)*(3/7)   (3/9)*(2/8)*(4/7)   (3/9) *(4/8)*(3/7)   (4/9)*(2/8)*(3/7)   (4/9)*(3/8)*(3/7)   total probability of getting a blue ball is sum of all above ways so p=(2/9)*(3/8)*(4/7)+(2/9)*(4/8)*(3/7)+(3/9)*(2/8)*(4/7)+(3/9) *(4/8)*(3/7)+(4/9)*(2/8)*(3/7)+(4/9)*(3/8)*(3/7)=1/3   so probability of ball to be blue is 1/3   Answer 1/3
<result>E</result>
A box contains 2 red, 3 black and 4 blue balls. 3 balls are randomly drawn from the box. If two balls are found to be different, what is the probability that the third ball is blue?