Question

# In a bag of M & M's, there are 80 M & Ms, with 11 red...

In a bag of M & M's, there are 80 M & Ms, with 11 red ones, 12 orange ones, 20 blue ones, 10 green ones, 18 yellow ones, and 9 brown ones. They are mixed up so that each candy piece is equally likely to be selected if we pick one.

( ALL ANSWERS ROUNDED TO THREE DECIMAL PLACES)

a) If we select one at random, what is the probability that it is yellow?
P(Yellow) ______________

b) If we select one at random, what is the probability that it is not orange?
P(Not Orange) _______________

c) If we select one at random, what is the probability that it is yellow or blue?
P(Yellow or blue) ____________

d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are orange?
P (first one is orange and second one is orange) ________________

e) If we select one, keep it, and then select a second one, what is the probability that the first one is yellow and the second one is brown?
P(First one is yellow and second one is brown) ________

Total number of M&M's = 80

a) Number of yellow M&M's = 18

Hence, P(yellow) = 18/80

= 0.225

b) Number of orange M&M's = 12

Hence, P(not orange) = 1 - P(orange)

= 1 - (12/80)

= 0.85

c) Number of yellow M&M's = 18

Number of blue M&M's = 20

Hence, P(yellow or blue) = 18/80 + 20/80

= 0.475

d) Number of orange M&M's = 12

Hence, P(first one is orange and second one is orange) =12/80 * 12/80

= 0.026

e) Number of yellow M&M's = 18

Number of brown M&M's = 9

Hence, P(First one is yellow and second one is brown) = (18/80) * (9/79)

= 0.025

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