There are only [tex]r[/tex] red counters and [tex]g[/tex] green counters in a bag. A counter is taken at random from the bag. The probability that the counter is green is [tex]\frac{4}{9}[/tex]. The counter is put back in the bag. Four more red counters and two more green counters are added to the bag. A counter is taken from the bag again. The probability that the counter is green is [tex]\frac{10}{23}[/tex].

Find the number of red counters and the number of green counters that were in the bag originally.

Sure, let's solve this probability problem step by step.

1. Understanding the initial setup:
- Let [tex]\( r \)[/tex] be the number of red counters and [tex]\( g \)[/tex] be the number of green counters originally in the bag.

2. First Condition:
- We are given that the probability of picking a green counter initially is [tex]\(\frac{4}{9}\)[/tex].
- This gives us the equation:
[tex]\[
\frac{g}{r + g} = \frac{4}{9}
\][/tex]
- Solving for [tex]\( g \)[/tex] in terms of [tex]\( r \)[/tex], we get:
[tex]\[
9g = 4(r + g)
\][/tex]
[tex]\[
9g = 4r + 4g
\][/tex]
[tex]\[
5g = 4r
\][/tex]
[tex]\[
g = \frac{4}{5}r
\][/tex]

3. Second Condition:
- After adding 4 more red counters and 2 more green counters, the number of red counters becomes [tex]\( r + 4 \)[/tex] and the number of green counters becomes [tex]\( g + 2 \)[/tex].
- The new probability of picking a green counter is [tex]\(\frac{10}{23}\)[/tex].
- This gives us the equation:
[tex]\[
\frac{g + 2}{r + 4 + g + 2} = \frac{10}{23}
\][/tex]
- Solve for this:
[tex]\[
\frac{g + 2}{r + g + 6} = \frac{10}{23}
\][/tex]
[tex]\[
23(g + 2) = 10(r + g + 6)
\][/tex]
[tex]\[
23g + 46 = 10r + 10g + 60
\][/tex]
[tex]\[
13g = 10r + 14
\][/tex]
[tex]\[
g = \frac{10}{13}r + \frac{14}{13}
\][/tex]

4. Solve the equations:
- Now, we have two equations:
- [tex]\( g = \frac{4}{5}r \)[/tex]
- [tex]\( g = \frac{10}{13}r + \frac{14}{13} \)[/tex]

- Set them equal to each other to solve for [tex]\( r \)[/tex]:
[tex]\[
\frac{4}{5}r = \frac{10}{13}r + \frac{14}{13}
\][/tex]

- Solve for [tex]\( r \)[/tex]:
[tex]\[
\frac{4}{5}r - \frac{10}{13}r = \frac{14}{13}
\][/tex]

- Get a common denominator (65):
[tex]\[
\frac{52}{65}r - \frac{50}{65}r = \frac{14}{13}
\][/tex]
[tex]\[
\frac{2}{65}r = \frac{14}{13}
\][/tex]
[tex]\[
r = \frac{14}{13} \times \frac{65}{2}
\][/tex]
[tex]\[
r = 35
\][/tex]

5. Finding [tex]\( g \)[/tex]:
- Substitute [tex]\( r = 35 \)[/tex] into [tex]\( g = \frac{4}{5}r \)[/tex]:
[tex]\[
g = \frac{4}{5} \times 35
\][/tex]
[tex]\[
g = 28
\][/tex]

Thus, the number of red counters originally in the bag is 35, and the number of green counters is 28.