Question

M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the distribution of colors for plain M&M candies is as follows. Color Purple Yellow Red Orange Green Blue Brown Percentage 23% 22% 19% 6% 6% 10% 14% Suppose you have a large bag of plain M&M candies and you choose one candy at random. (a) Find P(green candy or blue candy). Are these outcomes mutually exclusive? Why? Yes. Choosing a green and blue M&M is possible. No. Choosing a green and blue M&M is possible. Yes. Choosing a green and blue M&M is not possible. No. Choosing a green and blue M&M is not possible. (b) Find P(yellow candy or red candy). Are these outcomes mutually exclusive? Why? Yes. Choosing a yellow and red M&M is not possible. No. Choosing a yellow and red M&M is not possible. No. Choosing a yellow and red M&M is possible. Yes. Choosing a yellow and red M&M is possible. (c) Find P(not purple candy).

Answer 1

Answer:

Step-by-step explanation:

Hello!

The distribution of colors for plain M&M candies is:

Purple 23% (U)

Yellow 22% (Y)

Red 19% (R)

Orange 6% (O)

Green 6% (G)

Blue 10% (B)

Brown 14%. (W)

If you choose ONE candy at random:

a. Two events are mutually exclusive when the ocurrence of one of them prevents the other from happening in one random choise. In this case, when you take one candy and its color is, for example, green, then it cannot be blue, this means that the events "green" and "blue" are mutually exclusive.

P(G∪B) = P(G) + P(B) - P(G∩B) = 0.06 + 0.1 = 0.16

b) The events yellow and red are mutually exclusive (same explanation as in item a.)

P(Y∪R)= P(Y) + P(R) - P(Y∩R) = 0.22 + 0.19 = 0.41

c) The probability of the candy not being purple.

P(U')= 1 - P(U)= 1 - 0.23= 0.77

I hope you have a SUPER day!