Question 1191331
According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 
12% are red, 23% are blue, 23% are orange and 15% are green. 
You randomly select peanut M&M’s from an extra-large bag looking for a yellow candy. 
(Round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 12.34%.)

a) Compute the probability that the first yellow candy is the sixth M&M selected.

b) Compute the probability that the first yellow candy is the sixth or seventh M&M selected.

c) Compute the probability that the first yellow candy is among the first six M&M’s selected.

d) If every student in a large Statistics class selects peanut M&M’s at random until they get a yellow candy, 
on average how many M&M’s will the students need to select? (Round your answer to two decimal places.)
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Tooooooo many questions to answer all of them at a time.


In this post,  I will answer  (a),  (b)  and  (c),  only.


The major idea is to recognize that we have a Binomial distribution probability problem with the success 
individual trial probability of  0.15  for yellow and  (1-0.15) = 0.85  for not yellow.



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(a)  P(the first yellow candy is the sixth M&M selected) = 0.85^5*0.15 = 0.06656   (rounded).



(b)  P(the first yellow candy is the sixth or seventh M&M selected) = 0.85^5*0.15 + 0.85^6*0.15 = 0.1231     (rounded).



(c)  P(the first yellow candy is among the first six M&M’s selected) = 

         = is the same as (simply re-phrasing) =  

  =  P(exactly one yellow is among the first six M&M’s selected) = 

  =  {{{C[6]^1*0.85^5*0.15}}} = {{{6*0.85^5*0.15}}} = 0.3993.
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