Question

Peanut M & Ms
In a bag of M & M's, there are 80 M & Ms, with  10 red ones,  11 orange ones,  18 blue ones,  10 green ones,  16 yellow ones, and  15 brown ones. They are mixed up so that each candy piece is equally likely to be selected if we pick one.

(a) If we select one at random, what is the probability that it is red?
Round your answer to three decimal places.

​​​​​​(b) If we select one at random, what is the probability that it is not green?
Round your answer to three decimal places.

(c) If we select one at random, what is the probability that it is red or yellow?
Round your answer to three decimal places.

(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 green?
Round your answer to three decimal places.
P(first one is green and second one is green)=

​​​​​​(e) If we select one, keep it, and then select a second one, what is the probability that the first one is red and the second one is blue?
Round your answer to three decimal places.

P(first one is red and second one is blue)=

Answer 1

Total M&M = 80

a) P(red) = 10/80 = 0.125

b) P(not green) = 1 - P(green) = 1 - 10/80 = 70/80 = 0.875

c) P(Red or yellow) = (10+16)/80 = 0.325

d) P(green, green) = (10/80)*(10/80) = 1/64 = 0.016 (rounded to 3 decimal places)

e) P(red, blue) = (10/80)*(18/79) = 9/316 = 0.028 (rounded to 3 decimal places)

Please comment if any doubt. Thank you.