The only contents of a container are 4 blue disks and 8 green disks. If 3 disks are selected one after the other, and at random and without replacement from the container, what is the probability that 1 of the disks selected is blue, and 2 of the disks selected are green?
Choices.
A) 21/55
B) 28/55
C) 34/55
D) 5/8
E) 139/220
Prob of selecting one blue and two green is (4*8*7/12*11*10)   one blue and two green can be arranged in 3!/2! ways   hence the prob becomes 3*4*8*7/12*11*10=28/55   Correct answer is 28/55
<result>B</result>
The only contents of a container are 4 blue disks and 8 green disks. If 3 disks are selected one after the other, and at random and without replacement from the container, what is the probability that 1 of the disks selected is blue, and 2 of the disks selected are green?