Answer: [tex]\dfrac{13}{2660}[/tex]
Step-by-step explanation:
Given : A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow.
Total candies = 12+12+7+13+3+10=57
Three candies are pulled from the bag in succession, without replacement.
Number of combinations of selecting r things out of n things = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]
Number of combinations of selecting 3 things out of 57 things
= [tex]^{57}C_3=\dfrac{57!}{3!(57-3)!}\\\\=\dfrac{57\times56\times55\times54!}{(6)(4)!}\\\\=57\times56\times55=175560[/tex]
i.e. Total outcomes = 175560
Number of combination of selecting 2 blue candies out of 12
=[tex]^{12}C_2=\dfrac{12!}{2!(12-2)!}\\\\=\dfrac{12\times11\times10!}{2\times10!}=66[/tex]
Number of combination of selecting 1 brown candies out of 13
=[tex]^{13}C_1=\dfrac{13!}{1!(13-1)!}\\\\=\dfrac{13\times12!}{1\times12!}=13[/tex]
Favorable outcomes = No. of combination of selecting 2 blue candies x No. of combination of selecting 1 brown candies
= 66 x 13 = 858
The probability that the first two candies drawn are blue and the third is brown =[tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
[tex]=\dfrac{858}{175560}\\\\=\dfrac{13}{2660}[/tex]
