Answer :
Let's solve the problem step by step:
1. Understanding the Probabilities:
- Initially, the probability of picking a green counter is given as [tex]\(\frac{4}{9}\)[/tex]. This means if we let [tex]\(r\)[/tex] be the number of red counters and [tex]\(g\)[/tex] be the number of green counters, the probability equation can be written as:
[tex]\[
\frac{g}{r + g} = \frac{4}{9}
\][/tex]
2. Equation from Initial Probability:
- Multiply both sides of the equation by [tex]\(r + g\)[/tex] to eliminate the fraction:
[tex]\[
g = \frac{4}{9}(r + g)
\][/tex]
- Simplify to get:
[tex]\[
9g = 4r + 4g
\][/tex]
- Rearrange to form equation 1:
[tex]\[
5g = 4r \quad \text{(Equation 1)}
\][/tex]
3. Understanding the Second Scenario:
- After adding 4 red and 2 green counters, the new number of red counters is [tex]\(r + 4\)[/tex], and green counters is [tex]\(g + 2\)[/tex].
- The probability of picking a green counter now is given as [tex]\(\frac{10}{23}\)[/tex], so:
[tex]\[
\frac{g + 2}{r + 4 + g + 2} = \frac{10}{23}
\][/tex]
4. Equation from Second Probability:
- Simplifying this, we get:
[tex]\[
(g + 2) = \frac{10}{23}(r + 4 + g + 2)
\][/tex]
- Multiply both sides by 23:
[tex]\[
23(g + 2) = 10(r + 6 + g)
\][/tex]
- Expand to get:
[tex]\[
23g + 46 = 10r + 60 + 10g
\][/tex]
- Rearrange to form equation 2:
[tex]\[
13g + 46 = 10r + 60
\][/tex]
[tex]\[
13g = 10r + 14 \quad \text{(Equation 2)}
\][/tex]
5. Solve the System of Equations:
- We have two equations:
1. [tex]\(5g = 4r\)[/tex]
2. [tex]\(13g = 10r + 14\)[/tex]
- From the first equation, express [tex]\(r\)[/tex] in terms of [tex]\(g\)[/tex]:
[tex]\[
r = \frac{5g}{4}
\][/tex]
- Substitute into the second equation:
[tex]\[
13g = 10\left(\frac{5g}{4}\right) + 14
\][/tex]
- Simplifying gives:
[tex]\[
13g = \frac{50g}{4} + 14
\][/tex]
[tex]\[
13g = 12.5g + 14
\][/tex]
[tex]\[
0.5g = 14
\][/tex]
[tex]\[
g = 28
\][/tex]
- Substitute [tex]\(g = 28\)[/tex] back into the expression for [tex]\(r\)[/tex]:
[tex]\[
r = \frac{5 \times 28}{4} = 35
\][/tex]
Therefore, there were originally 35 red and 28 green counters in the bag.
1. Understanding the Probabilities:
- Initially, the probability of picking a green counter is given as [tex]\(\frac{4}{9}\)[/tex]. This means if we let [tex]\(r\)[/tex] be the number of red counters and [tex]\(g\)[/tex] be the number of green counters, the probability equation can be written as:
[tex]\[
\frac{g}{r + g} = \frac{4}{9}
\][/tex]
2. Equation from Initial Probability:
- Multiply both sides of the equation by [tex]\(r + g\)[/tex] to eliminate the fraction:
[tex]\[
g = \frac{4}{9}(r + g)
\][/tex]
- Simplify to get:
[tex]\[
9g = 4r + 4g
\][/tex]
- Rearrange to form equation 1:
[tex]\[
5g = 4r \quad \text{(Equation 1)}
\][/tex]
3. Understanding the Second Scenario:
- After adding 4 red and 2 green counters, the new number of red counters is [tex]\(r + 4\)[/tex], and green counters is [tex]\(g + 2\)[/tex].
- The probability of picking a green counter now is given as [tex]\(\frac{10}{23}\)[/tex], so:
[tex]\[
\frac{g + 2}{r + 4 + g + 2} = \frac{10}{23}
\][/tex]
4. Equation from Second Probability:
- Simplifying this, we get:
[tex]\[
(g + 2) = \frac{10}{23}(r + 4 + g + 2)
\][/tex]
- Multiply both sides by 23:
[tex]\[
23(g + 2) = 10(r + 6 + g)
\][/tex]
- Expand to get:
[tex]\[
23g + 46 = 10r + 60 + 10g
\][/tex]
- Rearrange to form equation 2:
[tex]\[
13g + 46 = 10r + 60
\][/tex]
[tex]\[
13g = 10r + 14 \quad \text{(Equation 2)}
\][/tex]
5. Solve the System of Equations:
- We have two equations:
1. [tex]\(5g = 4r\)[/tex]
2. [tex]\(13g = 10r + 14\)[/tex]
- From the first equation, express [tex]\(r\)[/tex] in terms of [tex]\(g\)[/tex]:
[tex]\[
r = \frac{5g}{4}
\][/tex]
- Substitute into the second equation:
[tex]\[
13g = 10\left(\frac{5g}{4}\right) + 14
\][/tex]
- Simplifying gives:
[tex]\[
13g = \frac{50g}{4} + 14
\][/tex]
[tex]\[
13g = 12.5g + 14
\][/tex]
[tex]\[
0.5g = 14
\][/tex]
[tex]\[
g = 28
\][/tex]
- Substitute [tex]\(g = 28\)[/tex] back into the expression for [tex]\(r\)[/tex]:
[tex]\[
r = \frac{5 \times 28}{4} = 35
\][/tex]
Therefore, there were originally 35 red and 28 green counters in the bag.
