Compute the probability .9

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 peanut M&M’s from an extra-large bag looking for a blue candy. (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 the first blue candy is the fourth or fifth M&M selected.

Compute the probability that the first blue candy is among the first four M&M’s selected.

Answer 1

Since the candy bag is "extra large", we can considering the drawing of each M and M independent, drawing of a blue candy a "success", not drawing a "failures", the probability of a drawing a blue candy constant at 0.23.

To get 1st blue candy on 4th try, the first 3 tries have to be failures (probability of 0.77), so the probability of getting the first blue candy on the 4th try is (0.77)^3 * (0.23). Similarly, the probability of getting the first blue candy on the 5th try is (0.77)^4 * (0.23). The probability of getting the first blue candy on the 4th or 5th dry is the sum of those probabilities.

In determining the probability of getting the first blue candy in any of the first 4 tries, having a fixed number of tries allows us to use the binomial distribution to determine the probability: C(4,1)* (0.23) * (0.77)^3, where C(4,1) = 4!/3!1!