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 yellow.



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



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



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



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



Let's break down each question systematically.

### 1. **Probability that a randomly selected peanut M&M is *not* yellow:**

The probability that a peanut M&M is yellow is given as \( P(\text{Yellow}) = 0.15 \). So, the probability that it is **not yellow** is:

\[
P(\text{Not Yellow}) = 1 - P(\text{Yellow}) = 1 - 0.15 = 0.85
\]

Thus, the probability that a peanut M&M is not yellow is:

\[
\boxed{0.850}
\]

---

### 2. **Probability that a randomly selected peanut M&M is *yellow* or *orange*:**

The probability that a peanut M&M is yellow is \( P(\text{Yellow}) = 0.15 \), and the probability that a peanut M&M is orange is \( P(\text{Orange}) = 0.23 \).

Since the two events are mutually exclusive (they can't happen at the same time), we can simply add the probabilities:

\[
P(\text{Yellow or Orange}) = P(\text{Yellow}) + P(\text{Orange}) = 0.15 + 0.23 = 0.38
\]

Thus, the probability that a peanut M&M is yellow or orange is:

\[
\boxed{0.380}
\]

---

### 3. **Probability that three randomly selected peanut M&M’s are all green:**

The probability that a peanut M&M is green is \( P(\text{Green}) = 0.15 \).

If we select three M&M's independently, the probability that all three are green is:

\[
P(\text{All Green}) = P(\text{Green}) \times P(\text{Green}) \times P(\text{Green}) = 0.15^3 = 0.15 \times 0.15 \times 0.15 = 0.003375
\]

Thus, the probability that three peanut M&M’s are all green is:

\[
\boxed{0.003}
\]

---

### 4. **Probability that neither of two randomly selected peanut M&M’s are yellow:**

The probability that a peanut M&M is **not yellow** is \( P(\text{Not Yellow}) = 1 - P(\text{Yellow}) = 0.85 \).

For two independently selected M&M’s, the probability that neither of them is yellow is:

\[
P(\text{Neither Yellow}) = P(\text{Not Yellow}) \times P(\text{Not Yellow}) = 0.85 \times 0.85 = 0.7225
\]

Thus, the probability that neither of two peanut M&M’s are yellow is:

\[
\boxed{0.723}
\]

---

### 5. **Probability that at least one of two randomly selected peanut M&M’s is yellow:**

To find the probability that **at least one** of two M&M's is yellow, we can use the complement rule. The complement of "at least one is yellow" is "neither is yellow," which we've already computed as \( P(\text{Neither Yellow}) = 0.723 \).

Thus, the probability that **at least one** is yellow is:

\[
P(\text{At Least One Yellow}) = 1 - P(\text{Neither Yellow}) = 1 - 0.723 = 0.277
\]

Thus, the probability that at least one of two peanut M&M’s is yellow is:

\[
\boxed{0.277}
\]

---

Do you want further details or have any questions?

### Related Questions:
1. What is the probability that a randomly selected peanut M&M is red or blue?
2. What is the probability that two randomly selected peanut M&M’s are both orange?
3. What is the probability that at least one of three peanut M&M's is green?
4. What is the probability that none of three peanut M&M’s is blue?
5. What is the probability that a randomly selected peanut M&M is brown or green?

### Tip:
When dealing with probabilities of "at least one," it's often easier to compute the complement (the probability that none occurs) and subtract from 1.

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 yellow.

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

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

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

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

Learn how to solve probability problems using peanut M&M color distribution. Compute probabilities for colors not being yellow, either yellow or orange, and three consecutive green M&Ms, plus probabilities for selecting two M&Ms without yellow. Suitable for Grades 9-12.

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