Question

According to a recent study, 21% of peanut M&M’s are brown, 5% are yellow, 16% are red, 11% are blue, 32% are orange, and 16% are green. Assume these proportions are correct and suppose you randomly select five peanut M&M’s from an extra-large bag of the candies. Calculate the following probablities. Also calculate the mean and standard deviation of the distribution. Round all solutions to four decimal places, if necessary.

Compute the probability that exactly four of the five M&M’s are green.

P(x=4)=

Compute the probability that three or four of the five M&M’s are green.

P(x=3 or x=4)=

Compute the probability that at most four of the five M&M’s are green.

P(x≤4)=

Compute the probability that at least four of the five M&M’s are green.

P(x≥4)=

If you repeatedly select random samples of five peanut M&M’s, on average how many do you expect to be green?

μ= green M&M’s

With what standard deviation?

σ= green M&M’s

Answer 1

Solution :

=> From the given information,

=> P(M&M’s are green) = p = 0.16 , n = 5

=> q = 1 - p = 0.84

=> From binomial distribution, P(X = r) = nCr*p^r*q^(n-r)

=> P(x = 4) = 5C4*0.16^4*0.84^1

= 0.0028

=> P(x = 3 or x = 4) = P(x = 3) + P(x = 4)

= 5C3*0.16^3*0.84^2 + 5C4*0.16^4*0.84^1

= 0.0289 + 0.0028

= 0.0317

=> P(x >= 4) = P(x = 4) + P(x = 5)

= 5C4*0.16^4*0.84^1 + 5C5*0.16^5*0.84^0

= 0.0028 + 0.0001

= 0.0029

=> Mean μ = n*p

= 5*0.16
  
= 0.8000

=> Standard deviation σ = sqrt(n*p*q)

= sqrt(5*0.16*0.84)

= 0.8198