Question

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

Answer 1

Answer: B.  $\dfrac{28}{55}$ .

Step-by-step explanation:

Given : Number of blue disks =4

Number of green disks = 8

Total disks = 12

Total number of combinations of drawing any 3 disks from 12 = $^{12}C_3$

Number of combinations of drawing 1 blue and 2 green disks = $^{4}C_1\times^{8}C_2$

Now , the probability that 1 of the disks selected is blue, and 2 of the disks selected are green will be :

$\dfrac{^{4}C_1\times^{8}C_2}{^{12}C_3}\\\\=\dfrac{4\times\dfrac{8!}{2!6!}}{\dfrac{12!}{3!9!}}\\\\=\dfrac{4\times28}{220}\\\\=\dfrac{28}{55}$

Hence, the correct answer is B.  $\dfrac{28}{55}$ .