Questions: According to Masterfoods, the company that manufactures MMs, 12% of peanut MMs 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 MM is not red. Compute the probability that a randomly selected peanut MM is brown or red. Compute the probability that three randomly selected peanut MMs are all yellow (assume they are drawn at the same time). If you randomly select three peanut MMs, compute that probability that none of them are blue. If you randomly select three peanut MMs, compute that probability that at least one of them is blue.

Using the formula for combined probabilities, $P(\text{brown or red}) = 0.12 + 0.12 = 0.24.$

Since each selection is independent, the probability that all 1 selections are brown or red is $P(\text{brown or red})^{num_selections} = 0.24^{num_selections} = 0.24.$

Using the probability of selecting a yellow, $P(\text{yellow}) = 0.15,$ for 3 independent selections. The probability that all 3 selections are yellow is $P(\text{yellow})^{num_selections} = 0.15^{num_selections} = 0.003.$

Using the formula for the complement probability, $P(\text{not blue}) = 1 - P(\text{blue}) = 1 - 0.23 = 0.77.$

Since each selection is independent, the probability that all 3 selections are not blue is $P(\text{not blue})^{num_selections} = 0.77^{num_selections} = 0.457.$

First, calculate the probability of not selecting a blue in one selection, $P(\text{not blue}) = 1 - P(\text{blue}) = 1 - 0.23 = 0.77.$

The probability of selecting at least one blue among 3 selections is $1 - P(\text{all not blue}) = 1 - (0.77)^{num_selections} = 0.543.$