Question

7. A bag of M&Ms contains the following distribution of colors:

9 blue
6 orange
5 brown
5 green
4 red
3 yellow
Three M&Ms are randomly selected without replacement. Find the probabilities of the following events.
a. All three are blue.
b. The first one selected is blue, the second one selected is orange, and the third one selected is red.
c. The first two selected are red, and the third one selected is yellow.

Answer 1

a) 21/1240 = 0.017

b) 9/1240 = 7.26*10^-3

c) 3/2480 = 1.21*10^-3

Explanation:

There are total 32 M&Ms in the bag

a) Probability that all three are blue M&Ms (without replacement)

(9/32)*(8*31)*(7/30) = 21/1240 = 0.017

b) Probability that the first one selected is blue, the second one selected is orange, and the third one selected is red

(9/32)*(6/31)*(4/30) = 9/1240 = 7.26*10^-3

c) Probability that the first two selected are red, and the third one selected is yellow

(4/32)*(3/31)*(3/30) = 3/2480 = 1.21*10^-3