Answer: B. [tex]\dfrac{28}{55}[/tex] .
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 = [tex]^{12}C_3[/tex]
Number of combinations of drawing 1 blue and 2 green disks = [tex]^{4}C_1\times^{8}C_2[/tex]
Now , the probability that 1 of the disks selected is blue, and 2 of the disks selected are green will be :
[tex]\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}[/tex]
Hence, the correct answer is B. [tex]\dfrac{28}{55}[/tex] .
