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 green.

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

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

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

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

Let's solve the problems one by one.

### 1. Compute the probability that a randomly selected peanut M&M is **not green**.

The probability that an M&M is green is given as \( P(\text{Green}) = 0.15 \).

So, the probability that an M&M is **not green** is:
\[
P(\text{Not Green}) = 1 - P(\text{Green}) = 1 - 0.15 = 0.85
\]

### 2. Compute the probability that a randomly selected peanut M&M is **yellow or blue**.

The probability that an M&M is yellow is \( P(\text{Yellow}) = 0.15 \), and the probability that it is blue is \( P(\text{Blue}) = 0.23 \).

Since these are mutually exclusive events (i.e., the M&M cannot be both yellow and blue at the same time), the probability that it is yellow **or** blue is:
\[
P(\text{Yellow or Blue}) = P(\text{Yellow}) + P(\text{Blue}) = 0.15 + 0.23 = 0.38
\]

### 3. Compute the probability that **two randomly selected peanut M&M’s are both red**.

The probability that a single M&M is red is \( P(\text{Red}) = 0.12 \).

Since the selections are independent, the probability that **both** M&M’s are red is:
\[
P(\text{Both Red}) = P(\text{Red}) \times P(\text{Red}) = 0.12 \times 0.12 = 0.0144
\]

### 4. Compute the probability that **neither of two randomly selected peanut M&M’s are red**.

The probability that a single M&M is **not red** is:
\[
P(\text{Not Red}) = 1 - P(\text{Red}) = 1 - 0.12 = 0.88
\]
The probability that neither of the two M&M’s are red (i.e., both are not red) is:
\[
P(\text{Neither Red}) = P(\text{Not Red}) \times P(\text{Not Red}) = 0.88 \times 0.88 = 0.7744
\]

### 5. Compute the probability that **at least one of two randomly selected peanut M&M’s is red**.

The probability of **at least one being red** is the complement of **neither being red**. So, we subtract the probability that neither are red from 1:
\[
P(\text{At least one red}) = 1 - P(\text{Neither Red}) = 1 - 0.7744 = 0.2256
\]

### Summary of Answers:

1. \( P(\text{Not Green}) = 0.850 \)
2. \( P(\text{Yellow or Blue}) = 0.380 \)
3. \( P(\text{Both Red}) = 0.014 \)
4. \( P(\text{Neither Red}) = 0.774 \)
5. \( P(\text{At least one Red}) = 0.226 \)

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Would you like more details on any of the steps or further clarification?

Here are five related questions to extend the problem:
1. What is the probability that a randomly selected M&M is either brown or orange?
2. If you pick 3 M&M's, what is the probability that all are green?
3. What is the probability that none of 3 randomly selected M&M's are blue?
4. If you pick 3 M&M's, what is the probability that at least one is yellow?
5. What is the probability that at least one of two randomly selected M&M's is either red or blue?

**Tip**: When dealing with "at least one" probability questions, always consider using the complement rule (i.e., the probability of the event not happening) to simplify your calculations.

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 green. Compute the probability that a randomly selected peanut M&M is yellow or blue. Compute the probability that two randomly selected peanut M&M’s are both red. If you randomly select two peanut M&M’s, compute that probability that neither of them are red. If you randomly select two peanut M&M’s, compute that probability that at least one of them is red.

Explore the probabilities of selecting colored peanut M&M’s, including green, yellow, blue, and red. This problem demonstrates core probability concepts such as the complement rule and independent events. Suitable for students in Grades 9-12.

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