Question

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. Three candies are pulled from the bag in succession, without replacement. What is the probability that the first two candies drawn are blue and the third is brown?

Answer 1

Answer:
`(13)/(2660)`

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 =
`^nC_r=(n!)/(r!(n-r)!)`

Number of combinations of selecting 3 things out of 57 things

=
`^(57)C_3=(57!)/(3!(57-3)!)\\\\=(57*56*55*54!)/((6)(4)!)\\\\=57*56*55=175560`

i.e. Total outcomes = 175560

Number of combination of selecting 2 blue candies out of 12

=
`^(12)C_2=(12!)/(2!(12-2)!)\\\\=(12*11*10!)/(2*10!)=66`

Number of combination of selecting 1 brown candies out of 13

=
`^(13)C_1=(13!)/(1!(13-1)!)\\\\=(13*12!)/(1*12!)=13`

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 =
`\frac{\text{Favorable outcomes}}{\text{Total outcomes}}`

`=(858)/(175560)\\\\=(13)/(2660)`