Chapter P, Section 1, Exercise 036 More Peanut M & Ms In a bag of M & M\'s, ther

Question

Chapter P, Section 1, Exercise 036 More Peanut M & Ms In a bag of M & M's, there are 80 M & Ms, with 12 red ones, 13 orange ones, 20 blue ones, 10 green ones, 18 yellow ones, and 7 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 green? Round your answer to three decimal places. P(green) = the absolute tolerance is +/-0.005 (b) If we select one at random, what is the probability that it is not red? Round your answer to three decimal places. Pinot red) = the absolute tolerance is +/-0.005

Explanation / Answer

a) P( green ) = Total number of green ones / Total number of M&Ms in the bag = 10/80 = 0.125

Therefore 0.125 is the required probability here.

b) P( not red ) = Total number of non red ones / Total number of M&Ms in the bag = (80-12) /80 = 0.850

Therefore 0.850 is the required probability here.

c) P( green or orange) = Total number of green + orange ones / Total number of M&Ms in the bag = (10+13) /80 = 0.2875

Therefore 0.288 is the required probability here.

d) P( first one is red and second one is red ) = P( red )P(red ) = (12/80)*(12/80) = 0.0225

Therefore 0.023 is the required probability here.

e) Probability that the first is green and second is orange

= P( Green ) P( orange given that first is green )

= (10/80)*(13 / 79 )

Note that we used 79 here because after one has been drawn we are left with 79 after that

= 0.021

Therefore 0.021 is the required probability here.

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