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------------------------------------------------ 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 put in the bag. A counter is taken from the bag. 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.

Your final line must say: ... red and ... green counters.

To solve this problem, we will set up two equations based on the probabilities given and solve for the number of red ([tex]$r$[/tex]) and green ([tex]$g$[/tex]) counters originally in the bag.

### Step 1: Express the first probability

We know that the probability of picking a green counter from the original set is [tex]$\frac{4}{9}$[/tex].

[tex]\[
\frac{g}{r + g} = \frac{4}{9}
\][/tex]

This equation tells us that the number of green counters [tex]$g$[/tex] divided by the total number of counters [tex]$r + g$[/tex] equals [tex]$\frac{4}{9}$[/tex].

### Step 2: Express the second probability

After adding 4 red counters and 2 green counters, the total number of red counters becomes [tex]$r + 4$[/tex] and the total number of green counters becomes [tex]$g + 2$[/tex]. The total number of counters is now [tex]$(r + 4) + (g + 2)$[/tex] or [tex]$r + g + 6$[/tex].

We know the probability of picking a green counter now is [tex]$\frac{10}{23}$[/tex].

[tex]\[
\frac{g + 2}{r + g + 6} = \frac{10}{23}
\][/tex]

### Step 3: Solve the system of equations

Now we have a system of two equations:

1. [tex]$\frac{g}{r + g} = \frac{4}{9}$[/tex]
2. [tex]$\frac{g + 2}{r + g + 6} = \frac{10}{23}$[/tex]

To solve equation (1):
[tex]\[
g = \frac{4}{9}(r + g)
\][/tex]
[tex]\[
9g = 4(r + g)
\][/tex]
[tex]\[
9g = 4r + 4g
\][/tex]
[tex]\[
5g = 4r
\][/tex]
[tex]\[
r = \frac{5}{4}g
\][/tex]

Substitute [tex]$r = \frac{5}{4}g$[/tex] into equation (2):
[tex]\[
\frac{g + 2}{\frac{5}{4}g + g + 6} = \frac{10}{23}
\][/tex]
Simplify the denominator:
[tex]\[
\frac{g + 2}{\frac{9}{4}g + 6} = \frac{10}{23}
\][/tex]
Clear the fraction by multiplying both sides by [tex]$\frac{9}{4}g + 6$[/tex] and 23:
[tex]\[
23(g + 2) = 10(\frac{9}{4}g + 6)
\][/tex]
[tex]\[
23g + 46 = \frac{90}{4}g + 60
\][/tex]
Multiply through by 4 to clear fractions:
[tex]\[
92g + 184 = 90g + 240
\][/tex]
[tex]\[
92g - 90g = 240 - 184
\][/tex]
[tex]\[
2g = 56
\][/tex]
[tex]\[
g = 28
\][/tex]

Substitute back to find [tex]$r$[/tex]:
[tex]\[
r = \frac{5}{4}g = \frac{5}{4} \times 28 = 35
\][/tex]

### Conclusion
The number of red counters originally in the bag was 35, and the number of green counters was 28. Thus, the answer is:

35 red and 28 green counters.