Answer :
We start with the information that the bag contains only [tex]$r$[/tex] red counters and [tex]$g$[/tex] green counters, and initially the probability of drawing a green counter is
[tex]$$\frac{g}{r+g} = \frac{4}{9}.$$[/tex]
### Step 1. Establishing the Relationship Between [tex]$r$[/tex] and [tex]$g$[/tex]
Cross-multiply to obtain:
[tex]$$9g = 4(r+g).$$[/tex]
Expanding the right hand side:
[tex]$$9g = 4r + 4g.$$[/tex]
Subtract [tex]$4g$[/tex] from both sides:
[tex]$$9g - 4g = 4r,$$[/tex]
[tex]$$5g = 4r.$$[/tex]
Rearrange to express [tex]$r$[/tex] in terms of [tex]$g$[/tex]:
[tex]$$r = \frac{5}{4}g.$$[/tex]
### Step 2. Using the Changed Scenario
After the first draw (with replacement), 4 more red and 2 more green counters are added to the bag. This changes the totals to:
- Red counters: [tex]$r + 4$[/tex]
- Green counters: [tex]$g + 2$[/tex]
- Total counters: [tex]$r + g + 6$[/tex]
Now the probability of drawing a green counter becomes:
[tex]$$\frac{g+2}{r+g+6} = \frac{10}{23}.$$[/tex]
### Step 3. Substitute the Relation and Solve for [tex]$g$[/tex]
Substitute [tex]$r=\frac{5}{4}g$[/tex] into the total [tex]$r + g$[/tex]:
[tex]$$r + g = \frac{5}{4}g + g = \frac{9}{4}g.$$[/tex]
Thus, the equation becomes:
[tex]$$\frac{g+2}{\frac{9}{4}g + 6} = \frac{10}{23}.$$[/tex]
Cross multiply to eliminate the fraction:
[tex]$$23(g+2) = 10\left(\frac{9}{4}g + 6\right).$$[/tex]
Expand both sides:
[tex]$$23g + 46 = \frac{90}{4}g + 60.$$[/tex]
To simplify further, multiply the entire equation by 2 to eliminate the fraction:
[tex]$$2(23g + 46) = 2\left(\frac{90}{4}g + 60\right).$$[/tex]
This gives:
[tex]$$46g + 92 = \frac{90}{2}g + 120.$$[/tex]
Notice that [tex]$\frac{90}{2}g = 45g$[/tex], so:
[tex]$$46g + 92 = 45g + 120.$$[/tex]
Subtract [tex]$45g$[/tex] from both sides:
[tex]$$46g - 45g + 92 = 120,$$[/tex]
[tex]$$g + 92 = 120.$$[/tex]
Now subtract 92 from both sides to solve for [tex]$g$[/tex]:
[tex]$$g = 120 - 92,$$[/tex]
[tex]$$g = 28.$$[/tex]
### Step 4. Find [tex]$r$[/tex] Using the Relationship
Recall that
[tex]$$r = \frac{5}{4}g.$$[/tex]
Substitute [tex]$g = 28$[/tex]:
[tex]$$r = \frac{5}{4} \times 28 = 35.$$[/tex]
### Final Answer
There were originally
[tex]$$35$$[/tex] red counters and
[tex]$$28$$[/tex] green counters in the bag.
[tex]$$\frac{g}{r+g} = \frac{4}{9}.$$[/tex]
### Step 1. Establishing the Relationship Between [tex]$r$[/tex] and [tex]$g$[/tex]
Cross-multiply to obtain:
[tex]$$9g = 4(r+g).$$[/tex]
Expanding the right hand side:
[tex]$$9g = 4r + 4g.$$[/tex]
Subtract [tex]$4g$[/tex] from both sides:
[tex]$$9g - 4g = 4r,$$[/tex]
[tex]$$5g = 4r.$$[/tex]
Rearrange to express [tex]$r$[/tex] in terms of [tex]$g$[/tex]:
[tex]$$r = \frac{5}{4}g.$$[/tex]
### Step 2. Using the Changed Scenario
After the first draw (with replacement), 4 more red and 2 more green counters are added to the bag. This changes the totals to:
- Red counters: [tex]$r + 4$[/tex]
- Green counters: [tex]$g + 2$[/tex]
- Total counters: [tex]$r + g + 6$[/tex]
Now the probability of drawing a green counter becomes:
[tex]$$\frac{g+2}{r+g+6} = \frac{10}{23}.$$[/tex]
### Step 3. Substitute the Relation and Solve for [tex]$g$[/tex]
Substitute [tex]$r=\frac{5}{4}g$[/tex] into the total [tex]$r + g$[/tex]:
[tex]$$r + g = \frac{5}{4}g + g = \frac{9}{4}g.$$[/tex]
Thus, the equation becomes:
[tex]$$\frac{g+2}{\frac{9}{4}g + 6} = \frac{10}{23}.$$[/tex]
Cross multiply to eliminate the fraction:
[tex]$$23(g+2) = 10\left(\frac{9}{4}g + 6\right).$$[/tex]
Expand both sides:
[tex]$$23g + 46 = \frac{90}{4}g + 60.$$[/tex]
To simplify further, multiply the entire equation by 2 to eliminate the fraction:
[tex]$$2(23g + 46) = 2\left(\frac{90}{4}g + 60\right).$$[/tex]
This gives:
[tex]$$46g + 92 = \frac{90}{2}g + 120.$$[/tex]
Notice that [tex]$\frac{90}{2}g = 45g$[/tex], so:
[tex]$$46g + 92 = 45g + 120.$$[/tex]
Subtract [tex]$45g$[/tex] from both sides:
[tex]$$46g - 45g + 92 = 120,$$[/tex]
[tex]$$g + 92 = 120.$$[/tex]
Now subtract 92 from both sides to solve for [tex]$g$[/tex]:
[tex]$$g = 120 - 92,$$[/tex]
[tex]$$g = 28.$$[/tex]
### Step 4. Find [tex]$r$[/tex] Using the Relationship
Recall that
[tex]$$r = \frac{5}{4}g.$$[/tex]
Substitute [tex]$g = 28$[/tex]:
[tex]$$r = \frac{5}{4} \times 28 = 35.$$[/tex]
### Final Answer
There were originally
[tex]$$35$$[/tex] red counters and
[tex]$$28$$[/tex] green counters in the bag.
