According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's arebrown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. Yourandomly select peanut M&M's from an extra-large bag looking for a orange candy. (Round allprobabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444or 12.34%.)Compute the probability that the first orange candy is the sixth M&M selected.Compute the probability that the first orange candy is the sixth or seventh M&M selected.Compute the probability that the first orange candy is among the first six M&M's selected.If every student in a large Statistics class selects peanut M&M's at random until they get a orangecandy, on average how many M&M's will the students need to select? (Round your answer to twodecimal places.)orange M&M's

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Description: According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's arebrown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. Yourandomly select peanut M&M's from an extra-large bag looking for a orange ca

Answered: According to Masterfoods, the company…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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 orange candy. (Round all
probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444
or 12.34%.)
Compute the probability that the first orange candy is the sixth M&M selected.
Compute the probability that the first orange candy is the sixth or seventh M&M selected.
Compute the probability that the first orange candy is among the first six M&M's selected.
If every student in a large Statistics class selects peanut M&M's at random until they get a orange
candy, on average how many M&M's will the students need to select? (Round your answer to two
decimal places.)
orange M&M's

Expert Solution

Step 1

Given Information:

12% 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 the extra-large bag looking for a orange candy.

(a) To compute the probability that the first orange candy is the sixth M&M selected:

We are using geometric distribution.

Probability of selecting an orange candy is $p = 0.23$

Probability of not selecting an orange candy is $q = 1 - p = 1 - 0.23 = 0.77$

Formula:

$P (X = x) = p {(1 - p)}^{k - 1}$

Required probability is obtained as follows:

$\begin{array}{rcl} P (6) & = & 0.23 {(0.77)}^{6 - 1} \\ = & 0.23 {(0.77)}^{5} \\ = & 0.06225603561 \\ \approx & 0.0623 \end{array}$

Thus, the probability that the first orange candy is the sixth M&M selected is 0.0623

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