tessthetoast

12.08.2020 • 

Mathematics

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 22% 20% 23% 10% 6% 6% 13% 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. Yes. Choosing a green and blue M&M is not possible. No. 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? Yes. Choosing a yellow and red M&M is possible. No. Choosing a yellow and red M&M is not possible. Yes. Choosing a yellow and red M&M is not possible. No. Choosing a yellow and red M&M is possible. (c) Find P(not purple candy).

Ответ:

crybaby222psyc

30.01.2022

Mathematics

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A) 0.12. Yes. Choosing a green and blue M&M is possible

B) 0.43. Yes. Choosing a yellow and red M&M is possible

C) 0.78

Step-by-step explanation:

First of all, the summation of the distribution of all colours is;

Σ(all colors ) = 22% + 20% + 23% + 10% + 6% + 6% + 13% = 100%, or 1.

Thus;

a) P(green candy or blue candy) is;

P(GREEN ∪ BLUE) = P(G) + P(BL)

P(GREEN ∪ BLUE) = 6%+6%

P(GREEN ∪ BLUE) = 12% or 0.12

Now, due to the fact that we have to choose ONE candy and only ONE candy at random, then they are mutually exclusive: Yes. Choosing a green and blue M&M is possible

b)P(yellow candy or red candy is;

P(YELLOW ∪ RED) = P(Y) + P(R)

P(YELLOW ∪ RED) = 20% + 23% = 43% or 0.43

Yes. Choosing a yellow and red M&M is possible

c) P(NOT PURPLE)

the probability of having a purple is;

P(PURPLE) = 22% or 0.22

So, the Probability of NOT having a PURPLE is 1 - 0.22 = 0.78

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Ответ:

jojojolie6505

07.06.2022

Mathematics

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the expression is 2x - 3y + 8 you welcome

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