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 four of the five M&M’s are brown.

Answer 1

Answer:

0.0009

Step-by-step explanation:

n = 5; x = 4

p = 0.12; q = 1 – 0.12 = 0.88

The probability that exactly four of the five M&M's are brown is:

$_{n}C_{x} \times p^{x} \times q^{n-x}=\\\\_{5}C_{4} \times (0.12)^{4} \times (0.88)^{1}=\\\\5 \times 0.00020736 \times 0.88 =\\\\0.000912384 \approx 0.0009$