Question

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 five peanut M&M’s from an extra-large bag of the candies. (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 exactly two of the five M&M’s are yellow.

  

Compute the probability that two or three of the five M&M’s are yellow.

  

Compute the probability that at most two of the five M&M’s are yellow.

  

Compute the probability that at least two of the five M&M’s are yellow.

  

Answer 1

Sample size = n = 5 Probability of success which is M&M's are yellow = 0.15 X is the Number of M&M's are yellow ! Then, X ~ Bin (5, 0.15) P(x=x) 56x (0.15)*(0.8535-* , X=0,1,2,...,5 PlExactly two of the five M&M's are yellow) P (X=2) 56 (0.15² (0.85)3 0.1382

Pltwo or three of the five M&M's are yellow) = P(X= 2) + P(x= 3) (0.15)2(0.85)3 + t, (0.15)? (0.85) 0.1626 Plat most two of the five M&M's are yellow) P (832) P(x=0) + P(x= 1) + PLX=2) 25.10.15% (0.85€+ (0.15)(0-36)***((0.5)(0.65) 0.9734 Plat least two of the five M&M's are yellow) P ( X2) 1- P(X=0 ) - P(x=19. - 5C. (6.15) (086) -(085(0.5) 0.1648