sanjoo
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?
A. 21/55
B. 28/55
C. 34/55
D. 5/8
E. 139/220
1st Blue disk can be selected in 4C1 = 4 ways
1st Green disk can be selected in 8C1 = 8 ways
2nd Green disk (without replacement) can be selected in 7C1 = 7 ways
1 blue disk can be selected out of total 12 disks in 12C1 = 12 ways
1 green disk can be selected out of remaining 11 disks in 11C1 = 11 ways
2nd green disk can be selected out of remaining 10 disks in 10C1 = 10 ways
Probability = (4*8*7)/(12*11*10)
But, the order of ONE blue and TWO green could be arranged in 3!/2! ways.
So total probability = 3!/2! * (4*8*7)/(12*11*10)
= 28/55
Hence option B.
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