Find probability

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 six peanut M&M’s from an extra-large bag of the candies. (Round all probabilities below to four decimal places: your answer should look like 0.1234, not 0.1234444 or 12.34%.)

Compute the probability that exactly four of the six M&M’s are blue:

  

Compute the probability that four or five of the six M&M’s are blue:

  

Compute the probability that at most four of the six M&M’s are blue:

  

Compute the probability that at least four of the six M&M’s are blue:

  

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

   M&M’s

With what standard deviation?

   M&M’s

Answer 1

Br = brown M&M's; Y = yellow M&M's; R = red M&M's;

B = blue M&M's; O = orange M&M's; G = green M&M's

P(Br) = 0.12; P(Y) = 0.15; P(R) = 0.12;

P(B) = 0.23; P(O) = 0.23; P(G) = 0.15

n = 6 peanut M&M's in the large bag

P(exactly 4 blue M&M's) = C(6,4)*(0.12)⁴*(0.88)² = 0.002409

P(4 or 5 blue M&M's) = C(6,4)*(0.12)⁴*(0.88)² + C(6,5)*(0.12)⁵*(0.88)¹

= 0.002409 + 0.0001314

= 0.0025404

P(up to 4 blue M&M's) = 1 - P(more than 4 blue M&M's)

= 1 - [C(6,5)*(0.12)⁵*(0.88)¹ + C(6,6)*(0.12)⁶*(0.88)⁰]

= 1 - 0.0001344

≈ 0.9999

P(at least 4 blue M&M's) = C(6,4)*(0.12)⁴*(0.88)² + C(6,5)*(0.12)⁵*(0.88)¹ + C(6,6)*(0.12)⁶*(0.88)⁰

= 0.002409 + 0.0001314 + 0.000002986

≈ 0.002543

The expected value from the binomial distribution is np: (6)(0.12) = 0.72

The standard deviation from the binomial distribution is √npq: √(6)(0.12)(0.88) = 0.796