Let's solve each question step-by-step based on the contents of the image.

Information Provided

The bag of M&Ms contains:

• 3 red
• 7 green
• 8 blue
• 2 yellow

Total M&Ms = $3 + 7 + 8 + 2 = 20$

Question 1

If you already drew out a blue M&M, what's the probability of drawing out a yellow?

Since one blue M&M has already been drawn, there are now:

• Total M&Ms left = 19
• Yellow M&Ms = 2

Probability of drawing a yellow M&M: \frac{\text{Number of yellow M&Ms}}{\text{Total M&Ms left}} = \frac{2}{19}

Question 2

What's the probability of drawing a blue or green?

Initially, there are:

• Blue M&Ms = 8
• Green M&Ms = 7
• Total M&Ms = 20

Probability of drawing a blue or green M&M: \frac{\text{Number of blue M&Ms + Number of green M&Ms}}{\text{Total M&Ms}} = \frac{8 + 7}{20} = \frac{15}{20} = \frac{3}{4}

Question 3

If you already drew out a red M&M, what's the probability of drawing out another red?

After drawing one red M&M, the remaining count is:

• Total M&Ms left = 19
• Red M&Ms left = 2

Probability of drawing another red M&M: \frac{\text{Remaining red M&Ms}}{\text{Total M&Ms left}} = \frac{2}{19}

Summary of Answers:

1. $\frac{2}{19}$
2. $\frac{3}{4}$
3. $\frac{2}{19}$

Would you like further details on any of these calculations?

Related Questions:

1. What is the probability of drawing two blue M&Ms consecutively?
2. If two yellow M&Ms are drawn, what's the probability of then drawing a green M&M?
3. How would the probability change if there was one more red M&M in the bag?
4. What’s the probability of drawing a non-blue M&M?
5. What is the probability of drawing a red or yellow M&M?

Tip:

When calculating probabilities with dependent events (like drawing without replacement), always update the total count and the count of specific items after each draw.