To solve this problem, we need to determine the probability that exactly two out of the three marbles drawn are red when drawing without replacement.

Step 1: Total Number of Marbles

The total number of marbles is: $6 \text{ (red)} + 8 \text{ (white)} + 5 \text{ (blue)} = 19 \text{ marbles}.$

Step 2: Total Number of Ways to Draw 3 Marbles

The total number of ways to choose 3 marbles from 19 is given by: $\binom{19}{3} = \frac{19!}{3!(19-3)!} = \frac{19 \times 18 \times 17}{3 \times 2 \times 1} = 969.$

Step 3: Number of Ways to Draw Exactly Two Red Marbles

To get exactly two red marbles, we need to:

1. Choose 2 red marbles from the 6 available.
2. Choose 1 non-red marble from the remaining 13 marbles (8 white + 5 blue).

The number of ways to choose 2 red marbles from 6 is: $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15.$

The number of ways to choose 1 non-red marble from 13 is: $\binom{13}{1} = 13.$

Step 4: Total Number of Favorable Outcomes

The total number of favorable outcomes for drawing exactly two red marbles and one non-red marble is: $15 \times 13 = 195.$

Step 5: Calculate the Probability

The probability of drawing exactly two red marbles is: $P = \frac{195}{969} \approx 0.201.$

Thus, the probability that exactly two of the marbles drawn are red is approximately 0.201 or 20.1%.

Would you like more details or have any questions?

Related Questions

1. What is the probability that all three marbles drawn are red?
2. How does the probability change if you draw with replacement instead of without?
3. What is the probability of drawing at least one red marble?
4. How would the answer change if there were 10 red marbles instead of 6?
5. What is the probability of drawing one marble of each color?

Tip

When solving probability problems without replacement, consider the order of selection and how the sample size changes after each draw.