Question

In a bag of peanut \(M\) \& M's, there are \(80 \mathrm{M} \& \mathrm{Ms}\), with 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each candy piece is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is red? (b) If we select one at random, what is the probability that it is not blue? (c) If we select one at random, what is the probability that it is red or orange? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are blue? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is red and the second one is green?

Short answer

a) \( \frac{11}{80} \), b) \( \frac{60}{80} \), c) \( \frac{23}{80} \), d) \( \left(\frac{20}{80} \right)^2 \), e) \( \frac{11}{80} \times \frac{11}{79} \)

Step-by-step solution

Total number of M&Ms

Find the total number of M&Ms in the bag. This can be done by summing up the number of M&Ms of each color. So, total M&Ms are 80.

Find Probability for Question a

Calculate the probability of picking a red M&M which is given by the fraction of the number of red M&Ms over the total number of M&Ms, i.e., \( P(red) = \frac{11}{80} \).

Find Probability for Question b

Calculate the probability of not picking a blue M&M. This is done by subtracting the fraction of the number of blue M&Ms over the total number of M&Ms from 1. So, \( P(not blue) = 1 - \frac{20}{80} = \frac{60}{80} \)

Find Probability for Question c

Calculate the probability of picking either a red or an orange M&M. This is done by adding the probabilities of each, \( P(red or orange) = \frac{11}{80} + \frac{12}{80} = \frac{23}{80} \).

Find Probability for Question d

Calculate the probability that both the first and second M&Ms are blue. Since the M&M was placed back, the probabilities of the two events are independent, so simply square the probability of picking a blue M&M. So, \( P(blue and blue) = \left(\frac{20}{80} \right)^2 \).

Find Probability for Question e

Calculate the probability that the first M&M is red and the second one is green. The difference from the previous step is that, this time, the events are dependent (the first M&M is kept). So, we multiply the probability of the first one being red by the probability of the second one being green. This gives: \( P(red and green) = \frac{11}{80} \times \frac{11}{79} \), where the denominator for the second probability drops by one as there is one less M&M in the bag.