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. [Round your answers to three decimal places, for example: 0.123]

Compute the probability that a randomly selected peanut M&M is not orange.

Compute the probability that a randomly selected peanut M&M is green or yellow.

Compute the probability that two randomly selected peanut M&M’s are both brown.

If you randomly select two peanut M&M’s, compute that probability that neither of them are blue.

If you randomly select two peanut M&M’s, compute that probability that at least one of them is blue.

Answer 1

The probabilities for various selections of M&M's are computed, such as not orange (77%), green or yellow (30%), two brown (1.44%), neither blue (59.29%), and at least one blue (40.71%).

Step-by-step explanation:

The likelihood of not picking an orange M&M can be calculated by subtracting the orange M&M percentage from 100%. Thus, the probability of not orange M&M is 100% - 23% = 77%, or 0.770 when expressed as a decimal.

The probability of green or yellow M&M combines the probabilities of both colors, so that's 15% for green plus 15% for yellow, totaling 30%, or 0.300 in decimal form.

To find the probability of two brown M&Ms being selected consecutively, you multiply the chance of one brown M&M (12%) by itself because the selections are independent: 0.12 * 0.12 = 0.0144 or 1.44%.

If we want to pick two M&Ms and neither should be blue, we calculate the probability of not blue (1 - the probability of blue) for each pick and multiply them together: (1 - 0.23) * (1 - 0.23) = 0.77 * 0.77 = 0.5929 or 59.29%.

The probability that at least one M&M is blue when two are picked can be found by subtracting the probability that neither are blue from 1. From the previous calculation, we know the probability that neither is blue is 0.5929, so 1 - 0.5929 = 0.4071 or 40.71%.