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. (Round your answers to 4 decimal places where possible)

a. Compute the probability that a randomly selected peanut M&M is not red.



b. Compute the probability that a randomly selected peanut M&M is yellow or green.



c. Compute the probability that three randomly selected peanut M&M’s are all green.



d. If you randomly select five peanut M&M’s, compute that probability that none of them are orange.



e. If you randomly select five peanut M&M’s, compute that probability that at least one of them is orange.

Answer 1

Let,

Br: event that peanut M&M’s are brown.

Y: event that peanut M&M’s are yelow.

R: event that peanut M&M’s are red.

Bl: event that peanut M&M’s are blue.

O: event that peanut M&M’s are orange.

G: event that peanut M&M’s are green.

So, we have P(Br)= 12/100 P(Y) = 15/100 P(R) = 12/100

P(Bl) = 23/100 P(O) = 23/100 P(G) = 15/100

a) Probability that a randomly selected peanut M&M is not red = P(R') = 1 - P(R)

= 1 -12/100 = 88/100 = 0.88

  

b) Probability that a randomly selected peanut M&M is yellow or green = P(Y) + P(G)

  = 15/100 +15/100

= 30/100 = 0.30

c) We know, by binomial distribution P(X=x) = ⁿ C _x * p^x * (1-p)^(n-x)

Similarly,  probability that three randomly selected peanut M&M’s are all green is given by

here p = P(G) = 15/100 = 0.15 , n=3 and x= 3

P(G=3) = ³C₃ (0.15)³ * (1- 0.15) ⁰ = 1 * 0.003375 * 1 = 0.003375

d) Probability that out of five randomly select peanut M&M’s, none of them are orange, we can find it by binomial distribution like we did above.

Here p = P(O) = 23/100 = 0.23 , n = 5 & x=0

P(O=0) = ⁵C₅ (0.23)⁰ * (1- 0.23)⁵ = 1 * 1 * 0.2707 = 0.2707

e) Probability that out of five randomly selected peanut M&M’s, at least one of them is orange = P(O=1) + P(O=2) + P(O=3) + P(O=4) + P(O=5) = 1 - P(O=0)

= 1 - 0.2707 = 0.7293