Compute the 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 peanut M&M’s from an extra-large bag looking for a red 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 red candy is the fifth M&M selected.

Compute the probability that the first red candy is the fifth or sixth M&M selected.

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Compute the probability that the first red candy is among the first five M&M’s selected.

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Answer 1

Draw #1: Probability of "Red" = 12/100 and probability of "Not Red" = 88/100

Draw #2: Probability of "Red" = 12/99 and probability of "Not Red" = 87/99

Draw #3: Probability of "Red" = 12/98 and probability of "Not Red" = 86/98

Draw #4: Probability of "Red" = 12/97 and probability of "Not Red" = 85/97

Draw #5: Probability of "Red" = 12/96 and probability of "Not Red" = 84/96

So the probability that the first red candy is the fifth M&M selected is (88/100)(87/99)(86/98)(85/97)(12/96) = 0.0743

The probability that the first red candy is the sixth M&M selected is (88/100)(87/99)(86/98)(85/97)(84/96)(12/95) = 0.0078. So the probability that the first red candy is either the fifth or the sixth M&M selected is 0.0743 + 0.0078 = 0.0751

The probability that the first red candy is among the first five M&M’s selected is 12/100 + 12/99 + 12/98 + 12/97 + 12/96 = 0.6124