Question

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

Answer 1

Answer:

a) 21/1240 = 0.017

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

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

Step-by-step 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