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.
The question involves solving a system of equations derived from given probabilities for drawing a green counter from a bag before and after adding more counters. By setting up equations based on the probabilities, we determined that the bag initially contained 16 red and 20 green counters.
The problem here involves finding the original number of red and green counters in a bag based on provided probabilities, and then updating those numbers after adding more counters to the bag. Initially, the probability of drawing a green counter is 4/9. After adding 4 red and 2 green counters, the new probability of drawing a green counter becomes 10/23.
Let's denote the initial number of red counters as r and green counters as g. The probability of drawing a green counter is g / (r + g). From the first condition, we have g / (r + g) = 4/9. After adding counters, we have (g + 2) / (r + g + 6) = 10/23.
From the first condition, by cross-multiplying, we get 9g = 4r + 4g, which simplifies to 5g = 4r. From the second condition, we also cross-multiply and get 23g + 46 = 10r + 10g. Subbing the first equation into the second allows us to solve for the values of r and g. After solving these simultaneous equations, we find out there were 16 red and 20 green counters originally.
