Given the information, the original count of red and green counters in the bag were 9 and 7, respectively.
Let's denote the original number of red counters as ( r ) and the original number of green counters as ( g ).
Step 1: From the given information, we know that the probability of picking a green counter from the bag initially is [tex]\( \frac{g}{r+g} = \frac{4}{9} \).[/tex]
Step 2: After putting the counter back and adding more counters, the total number of red counters becomes ( r + 4 ) and the total number of green counters becomes ( g + 2 ).
Step 3: From the second scenario, the probability of picking a green counter from the bag becomes[tex]\( \frac{g+2}{r+4+g+2} = \frac{10}{23} \).[/tex]
Now, let's solve these equations step by step.
Step 1:
[tex]\[\frac{g}{r+g} = \frac{4}{9}\][/tex]
Cross multiply:
[tex]\[9g = 4(r+g)\][/tex]
[tex]\[9g = 4r + 4g\][/tex]
Move ( 4g ) to the left side:
9g - 4g = 4r
5g = 4r
Step 2:
After adding more counters, the total number of red counters becomes ( r + 4 ) and the total number of green counters becomes ( g + 2 ).
Step 3:
[tex]\[\frac{g+2}{r+4+g+2} = \frac{10}{23}\][/tex]
Cross multiply:
[tex]\[10(r+g+6) = 23(g+2)\][/tex]
[tex]\[10r + 10g + 60 = 23g + 46\][/tex]
Move terms around:
10r + 10g - 23g = 46 - 60
10r - 13g = -14
Now, we have two equations:
1. ( 5g = 4r )
2. ( 10r - 13g = -14 )
We can solve this system of equations to find the values of ( r ) and ( g ).
From equation 1, we can express ( r ) in terms of ( g ):
[tex]\[r = \frac{5g}{4}\][/tex]
Substitute this expression for ( r ) into equation 2:
[tex]\[10\left(\frac{5g}{4}\right) - 13g = -14\][/tex]
Simplify:
[tex]\[\frac{50g}{4} - 13g = -14\][/tex]
[tex]\[\frac{50g - 52g}{4} = -14\][/tex]
-2g = -14
Divide both sides by -2:
g = 7
Now, substitute the value of ( g ) back into equation 1 to find ( r ):
[tex]\[r = \frac{5 \times 7}{4} = \frac{35}{4} = 8\frac{3}{4}\][/tex]
Since the number of counters must be whole numbers, we see that ( r = 9 ).
So, there were originally 9 red counters and 7 green counters in the bag.
Final line: There were 9 red and 7 green counters.