SOLUTION: 5.2.10 The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s. a.) State the

Algebra ->  Probability-and-statistics -> SOLUTION: 5.2.10 The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s. a.) State the      Log On


   

Question 1185781: 5.2.10
The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s.
a.) State the random variable.
b.) Argue that this is a binomial experiment
Find the probability that
c.) Six M&M’s are brown.
d.) Twenty-five M&M’s are brown.
e.) All of the M&M’s are brown.
f.) Would it be unusual for a package to have only brown M&M’s? If this were to happen, what would you think is the reason?

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
n = 52
p = .14
q = 1 minus p = .86
p(x) = p^x * q^(n-x) * c(n,x)

p(6) = .14^6 * .86^46 * c(52,6) = 0.148745273


p(25) = .14^25 * .86^27 * c(52,25) = 3.66167 * 10^-9


p(52) = .14^52 * .86^0 * c(52,52) = 3.96879E * 10^-45

if you round your answers to 8 decimal places, those higher numbers become effectively 0.

i used excel to show all the probabilities.
the sum of those probabilities is equal to 1, as it should be.
note that the higher probbilitie are effectively equal to 0 when the results are rounded to 8 decimal digits.

here's the excel printout.





it would be unusual for a packet to have all brown m&m's.
the most likely cause would be that the other colors ran out and so there were only brown colors left.
this could be the result of machine failure or human failure.
maybe somebody forgot to load the other colors?