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 4 decimal places where possible)

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



b. Compute the probability that a randomly selected peanut M&M is red or brown.



c. Compute the probability that three randomly selected peanut M&M’s are all red.



d. If you randomly select three peanut M&M’s, compute that probability that none of them are green.



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

Answer 1

a. The probability of not selecting a red M&M is 88%. b. The exact probability of selecting a red or brown M&M is not provided. c. The probability of selecting three red M&M's is approximately 0.173%. d. The probability of not selecting any green M&M's out of three is approximately 61.41%. e. The probability of at least one green M&M out of three is approximately 38.59%.

Step-by-step explanation:

a. To find the probability that a randomly selected peanut M&M is not red, we need to subtract the probability of selecting a red M&M from 1. The probability of selecting a red M&M is given as 12%. So, the probability of not selecting a red M&M is 1 - 0.12 = 0.88, or 88%.

b. To find the probability that a randomly selected peanut M&M is red or brown, we need to add the probabilities of selecting a red M&M and a brown M&M. The probability of selecting a red M&M is 12% and the probability of selecting a brown M&M is not given. Without the probability for brown M&M, we cannot calculate the exact probability in this case.

c. To find the probability that three randomly selected peanut M&M's are all red, we need to multiply the probabilities of selecting a red M&M three times. The probability of selecting a red M&M is 12%, so the probability would be 0.12 * 0.12 * 0.12 = 0.001728, or approximately 0.173%.

d. To find the probability that none of the three randomly selected peanut M&M's are green, we need to calculate the probability of not selecting a green M&M for each selection. The probability of selecting a green M&M is 15%, so the probability of not selecting a green M&M for one selection is 1 - 0.15 = 0.85. Since there are three selections, we need to multiply this probability three times. The probability would be 0.85 * 0.85 * 0.85 = 0.614125, or approximately 61.41%.

e. To find the probability that at least one of the three randomly selected peanut M&M's is green, we can subtract the probability of none of them being green from 1. From part d, we know that the probability of none of the M&M's being green is 0.614125. So, the probability of at least one of them being green is 1 - 0.614125 = 0.385875, or approximately 38.59%.