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Let's break down and solve each part of the problem:

We are given a bag with 25 marbles in total, consisting of:
- 10 red marbles,
- 5 green marbles, and
- 10 blue marbles.

### (a) Probability of getting 3 green marbles when pulling out 5 marbles

This is a combination probability problem without replacement. We need to find the probability of drawing exactly 3 green marbles when pulling out 5 marbles.

The probability is calculated using combinations, specifically:

\[
P(\text{3 green}) = \frac{\binom{5}{3} \times \binom{20}{2}}{\binom{25}{5}}
\]

Where:
- \(\binom{5}{3}\) is the number of ways to choose 3 green marbles from 5 green marbles,
- \(\binom{20}{2}\) is the number of ways to choose the remaining 2 non-green marbles from the 20 non-green marbles,
- \(\binom{25}{5}\) is the total number of ways to choose 5 marbles from 25 marbles.

### (b) Probability of getting a green marble second, given you got a red marble first

This is a conditional probability problem.

- The probability of getting a red marble on the first draw is:
  \[
  P(\text{red first}) = \frac{10}{25}
  \]
- After drawing a red marble, there are now 24 marbles left, 5 of which are green. The probability of getting a green marble on the second draw is:
  \[
  P(\text{green second} \mid \text{red first}) = \frac{5}{24}
  \]

The overall probability is the product of these two probabilities:
\[
P(\text{green second and red first}) = \frac{10}{25} \times \frac{5}{24}
\]

### (c) Probability that the second marble is blue

Here, we first pull 2 marbles. We want to find the probability that the second marble is blue. This can occur in two ways:
1. The first marble is not blue, and the second is blue.
2. The first marble is blue, and the second marble is also blue.

Let’s break it down:
1. The probability that the first marble is **not** blue is \(\frac{15}{25}\) (since there are 15 marbles that are not blue), and the probability that the second is blue is \(\frac{10}{24}\).
2. The probability that the first marble is blue is \(\frac{10}{25}\), and the probability that the second is blue (after the first blue marble has been removed) is \(\frac{9}{24}\).

The total probability is:
\[
P(\text{second blue}) = \left(\frac{15}{25} \times \frac{10}{24}\right) + \left(\frac{10}{25} \times \frac{9}{24}\right)
\]

Let me calculate each of these probabilities.Here are the calculated probabilities:

### (a) Probability of getting 3 green marbles when pulling out 5 marbles:
\[
P(\text{3 green}) \approx 0.0358 \text{ or } 3.58\%
\]

### (b) Probability of getting a green marble second, given that you got a red marble first:
\[
P(\text{green second | red first}) \approx 0.0833 \text{ or } 8.33\%
\]

### (c) Probability that the second marble is blue:
\[
P(\text{second blue}) = 0.4 \text{ or } 40\%
\]

Would you like a more detailed explanation of any part of this solution? 

Here are five related questions to expand on this problem:
1. How would the probabilities change if there were 15 red marbles instead of 10?
2. What is the probability of drawing at least one green marble when pulling 5 marbles?
3. How do we calculate the expected number of green marbles in 5 draws?
4. What is the probability of pulling 2 blue marbles and 1 green marble in 3 draws?
5. How does the probability change if we put the marbles back after each draw?

**Tip**: When working with probability problems involving "without replacement," remember that each draw affects the probabilities of subsequent draws.

There is a bag containing 25 marbles of three different types: 10 red, 5 green, and 10 blue marbles.
a) You pull out 5 marbles. What is the probability of getting 3 green marbles?
b) You pull out two marbles. What is the probability of getting a green marble second, given you got a red marble first?
c) You pull out 2 marbles, what is the probability that the second marble is blue?

Explore how to calculate the probability of drawing 3 green marbles when pulling out 5 marbles, finding a green marble after drawing a red one, and determining the chances of getting a blue marble second. This detailed step-by-step solution applies key concepts in probability and combinations. Suitable for high school students in Grades 10-12.

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