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There are only [tex]$r$[/tex] red counters and [tex]$g$[/tex] green counters in a bag.

1. A counter is taken at random from the bag. The probability that the counter is green is [tex]$\frac{3}{7}$[/tex].

2. The counter is put back in the bag.

3. Two more red counters and three more green counters are added to the bag.

4. A counter is taken at random from the bag again. The probability that the counter is green is [tex]$\frac{6}{13}$[/tex].

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

Answer :

To solve this problem, we need to determine the number of red counters ([tex]\( r \)[/tex]) and green counters ([tex]\( g \)[/tex]) that were originally in the bag, based on the given probabilities.

1. Understanding the Problem:
- Initially, we have [tex]\( r \)[/tex] red counters and [tex]\( g \)[/tex] green counters.
- The probability of picking a green counter initially is given as [tex]\(\frac{3}{7}\)[/tex].

2. Setting up the First Equation:
[tex]\[
\text{Probability of Green initially} = \frac{g}{r + g} = \frac{3}{7}
\][/tex]
This equation implies:
[tex]\[
7g = 3(r + g)
\][/tex]
Simplifying, we get:
[tex]\[
7g = 3r + 3g \quad \Rightarrow \quad 4g = 3r \quad \text{(Equation 1)}
\][/tex]

3. After Adding New Counters:
- Two more red counters and three more green counters are added, so now we have [tex]\( r + 2 \)[/tex] red and [tex]\( g + 3 \)[/tex] green counters.
- The new total number of counters is [tex]\((r + 2) + (g + 3) = r + g + 5\)[/tex].
- The new probability of picking a green counter is [tex]\(\frac{6}{13}\)[/tex].

4. Setting up the Second Equation:
[tex]\[
\text{Probability of Green after adding counters} = \frac{g + 3}{r + g + 5} = \frac{6}{13}
\][/tex]
This equation implies:
[tex]\[
13(g + 3) = 6(r + g + 5)
\][/tex]
Expanding both sides, we get:
[tex]\[
13g + 39 = 6r + 6g + 30
\][/tex]
Simplifying further:
[tex]\[
7g = 6r - 9 \quad \text{(Equation 2)}
\][/tex]

5. Solving the Equations:
Now we have a system of two equations:
- [tex]\( 4g = 3r \)[/tex]
- [tex]\( 7g = 6r - 9 \)[/tex]

Solving this system of equations, we find:
- [tex]\( r = 12 \)[/tex]
- [tex]\( g = 9 \)[/tex]

6. Conclusion:
- Originally, there were 12 red counters and 9 green counters in the bag. These numbers satisfy both probability conditions given in the problem.

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