M&M plain candies come in various colors. According to the M&M/Mars Department o
ID: 3301839 • Letter: M
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
17%
22%
19%
6%
7%
6%
23%
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 not possible.
Yes. Choosing a green and blue M&M is not possible.
No. Choosing a green and blue M&M is possible.
(b) Find P(yellow candy or red candy).
???
Are these outcomes mutually exclusive? Why?
No. Choosing a yellow and red M&M is not possible.
Yes. Choosing a yellow and red M&M is possible.
No. Choosing a yellow and red M&M is possible.
Yes. Choosing a yellow and red M&M is not possible.
(c) Find P(not purple candy).
Color
Purple
Yellow
Red
Orange
Green
Blue
Brown
Percentage
17%
22%
19%
6%
7%
6%
23%
Explanation / Answer
a) P (green candy or blue candy) is computed as the probability that either a green candy or a blue candy is taken out
= P( green ) + P( blue )
= 0.07 + 0.06
= 0.13
Therefore 0.13 is the required probability here.
These outcomes are clearly mutually exclusive because we are taking out 1 M&M and we cannot have a green and a blue M&M in 1 draw.
Therefore Yes Choosing a green and blue M&M is not possible is the correct answer here.
b) P( yellow or red candy )
= P( yellow ) + P( red )
= 0.22 + 0.19
= 0.41
Therefore 0.41 is the required probability here.
These outcomes are clearly mutually exclusive because we are taking out 1 M&M and we cannot have a yellow and a red M&M in 1 draw.
Therefore Yes Choosing a yellow and red M&M is not possible is the correct answer here.
c) P( not purple only )
= 1 - P( purple )
= 1 - 0.17
= 0.83
Therefore 0.83 is the required probability here.
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