https://file.aiquickdraw.com/mathbot/file/1729745559851qfocw8bp.jpg

Let's solve the questions based on the given M&M distribution.

The total number of M&Ms in the bag is:
\[
5 \text{ (red)} + 6 \text{ (green)} + 8 \text{ (blue)} + 2 \text{ (yellow)} = 21
\]

### 1) Probability of picking a yellow M&M:
The number of yellow M&Ms is 2. Thus, the probability is:
\[
P(\text{yellow}) = \frac{2}{21}
\]

### 2) Probability of picking a blue or green M&M:
There are 8 blue and 6 green M&Ms, so the total for blue or green is:
\[
8 \text{ (blue)} + 6 \text{ (green)} = 14
\]
Thus, the probability is:
\[
P(\text{blue or green}) = \frac{14}{21} = \frac{2}{3}
\]

### 3) Probability of picking an orange M&M:
Since there are no orange M&Ms, the probability is:
\[
P(\text{orange}) = \frac{0}{21} = 0
\]

Let me know if you want any further details or have other questions!

Here are 5 related questions for further exploration:
1. What is the probability of picking a red M&M?
2. What is the probability of picking a non-yellow M&M?
3. What is the probability of picking either a red or a yellow M&M?
4. How does the probability change if we add 5 more yellow M&Ms to the bag?
5. What is the probability of picking two yellow M&Ms in a row without replacement?

**Tip:** When calculating probabilities, always start by determining the total possible outcomes and the favorable outcomes for each event.

A bag of M&Ms has 5 red, 6 green, 8 blue, and 2 yellow M&Ms. What is the probability of randomly picking: 1) a yellow? 2) a blue or green? 3) an orange?

Learn how to calculate the probability of picking a yellow, blue, green, or orange M&M from a bag with 5 red, 6 green, 8 blue, and 2 yellow M&Ms. This problem covers basic probability concepts. Suitable for students in Grades 5-8.

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