1. ## m&m probability

Hi can you look over my solution? I am having difficulty in C and D part(what do they mean by procedure?)
Thanks

1) According to company reports, the current mix of colors for M&M dark chocolate candies is: 20% yellow, 10% red, 15% orange, 10% blue, 25% green, with the rest being brown.
A. If a single piece is taken from a bag of 600 pieces, find the following probabilities:
a. P (red) = 0.10*600/600 = 60/600=1/10 = 0.1

b. P (brown) = 1- P (red) - P (Blue) - P (yellow) – P (green) – P (orange)
=1-(0.10+0.10+0.2+0.25+0.15)
= 1-0.8
= 0.2
c. P (blue or orange) = P (blue) + P(orange) = 0.10+0.15 = 0.25

d. P (NOT red) = 1-P (red) = 1- 0.1 = 0.9
e. P (NOT brown) = 1-P(brown) = 1-0.2 = 0.8
f. P (red and green) = 0 (as each m&m has single color, so we cannot find an m&m which is red and green)

B. If two pieces are taken from a 600 piece bag, find the following probabilities:
a. P (both are blue) b. P (the first is orange) = 90/600 = 0.15
= (60/600) * (59/599)

c. P (first is red and the second is green) = 60/600*150/599

C. How would the process in problem #1A change if we are sampling from a “fun-size” bag which only has 12 pieces?
Solution:
The probabilities of selecting a red, green, blue, orange or yellow m&m will remain the same, but the probabilities will not be realistic as m&ms are discrete and cannot be fractional, so if the probability of selecting a red m&m is 0.10 this implies that in a fun size packet there are 1.2 red m&ms.
D. How would the process in problem #1B change if we are sampling isfrom a “fun-size” bag which only has 12 pieces?

2. ## Re: m&m probability

Bump!........

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