Lock, Statistics: Unlocking the Power of Data, 2e Grace Period: 1 days left Regi

Question

Lock, Statistics: Unlocking the Power of Data, 2e Grace Period: 1 days left Register Now NEXT ASSIGNMENT RESOURCES Chapter P, Section 1, Exercise 036 More Peanur M & Ms Sum18-Assignment 3 In a bag of M & M's, there are 80 M & Ms, with 12 red ones, 13 orange ones, 18 blue ones, 11 green ones, 16 yellow ones, and 10 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 yellow? Round your answer to three decimal places. P(yellow) chatter P. Section Exercise 002 Chapter P, Section 1 the absolute tolerance is +/-0.005 Exercise 032 Chapter P, Section 1 (b) If we select one at random, what is the probability that it is not orange? Round your answer to three decimal places. hapter P, Sectio P(not orange)- Chapter P, Section 3, the absolute tolerance is+-0.005 Review Score Review Results by Study Objective (c) If we select one at random, what is the probability that it is yellow or blue? Round your answer to three decimal places. P(yellow or blue)- the absolute tolerance is +/-0.005 (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?

Explanation / Answer

Sol:

a) P( yellow ) = Total number of green ones / Total number of M&Ms in the bag = 16/80 = 0.2

Therefore 0.2 is the required probability here.

b) P( not orange ) = Total number of non red ones / Total number of M&Ms in the bag = (80-13) /80 = 0.8375

Therefore 0.8375 is the required probability here.

c) P( yellow or blue) = Total number of green + orange ones / Total number of M&Ms in the bag = (16+18) /80 = 0.425

Therefore 0.425 is the required probability here.

d) P( first orange is red and second orange is red ) = P( orange )P( orange ) = (13/80)*(13/80) = 0.02640

Therefore 0.02640 is the required probability here.

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