Answer :
To find the number of red counters and blue counters originally in the bag, we start by setting up the problem based on the probabilities given:
1. Initially, let's say there are [tex]\( r \)[/tex] red counters and [tex]\( b \)[/tex] blue counters in the bag.
2. The probability of picking a blue counter is given as [tex]\(\frac{4}{9}\)[/tex]. This translates to the equation for probability:
[tex]\[
\frac{b}{r + b} = \frac{4}{9}
\][/tex]
3. After returning the counter to the bag, 6 more red counters and 3 more blue counters are added.
4. Now, the probability of picking a blue counter changes to [tex]\(\frac{5}{12}\)[/tex] with the new total numbers of counters. This gives us another equation:
[tex]\[
\frac{b + 3}{(r + 6) + (b + 3)} = \frac{5}{12}
\][/tex]
From these two equations, we can solve for [tex]\( r \)[/tex] and [tex]\( b \)[/tex]:
- From the first equation:
[tex]\[
9b = 4(r + b)
\][/tex]
Simplifying, we get:
[tex]\[
9b = 4r + 4b
\][/tex]
[tex]\[
5b = 4r
\][/tex]
[tex]\[
r = \frac{5}{4}b \quad \text{(Equation 1)}
\][/tex]
- From the second equation:
[tex]\[
\frac{b + 3}{r + b + 9} = \frac{5}{12}
\][/tex]
Simplifying, we get:
[tex]\[
12(b + 3) = 5(r + b + 9)
\][/tex]
[tex]\[
12b + 36 = 5r + 5b + 45
\][/tex]
[tex]\[
7b - 5r = 9 \quad \text{(Equation 2)}
\][/tex]
Now, substitute [tex]\( r = \frac{5}{4}b \)[/tex] from Equation 1 into Equation 2:
- Substitute:
[tex]\[
7b - 5\left(\frac{5}{4}b\right) = 9
\][/tex]
[tex]\[
7b - \frac{25}{4}b = 9
\][/tex]
Multiply throughout by 4 to eliminate fractions:
[tex]\[
28b - 25b = 36
\][/tex]
[tex]\[
3b = 36
\][/tex]
[tex]\[
b = 12
\][/tex]
Now that we have [tex]\( b = 12 \)[/tex], use Equation 1 to find [tex]\( r \)[/tex]:
- [tex]\( r = \frac{5}{4} \times 12 = 15 \)[/tex]
Thus, the original number of red counters is 15, and the original number of blue counters is 12.
1. Initially, let's say there are [tex]\( r \)[/tex] red counters and [tex]\( b \)[/tex] blue counters in the bag.
2. The probability of picking a blue counter is given as [tex]\(\frac{4}{9}\)[/tex]. This translates to the equation for probability:
[tex]\[
\frac{b}{r + b} = \frac{4}{9}
\][/tex]
3. After returning the counter to the bag, 6 more red counters and 3 more blue counters are added.
4. Now, the probability of picking a blue counter changes to [tex]\(\frac{5}{12}\)[/tex] with the new total numbers of counters. This gives us another equation:
[tex]\[
\frac{b + 3}{(r + 6) + (b + 3)} = \frac{5}{12}
\][/tex]
From these two equations, we can solve for [tex]\( r \)[/tex] and [tex]\( b \)[/tex]:
- From the first equation:
[tex]\[
9b = 4(r + b)
\][/tex]
Simplifying, we get:
[tex]\[
9b = 4r + 4b
\][/tex]
[tex]\[
5b = 4r
\][/tex]
[tex]\[
r = \frac{5}{4}b \quad \text{(Equation 1)}
\][/tex]
- From the second equation:
[tex]\[
\frac{b + 3}{r + b + 9} = \frac{5}{12}
\][/tex]
Simplifying, we get:
[tex]\[
12(b + 3) = 5(r + b + 9)
\][/tex]
[tex]\[
12b + 36 = 5r + 5b + 45
\][/tex]
[tex]\[
7b - 5r = 9 \quad \text{(Equation 2)}
\][/tex]
Now, substitute [tex]\( r = \frac{5}{4}b \)[/tex] from Equation 1 into Equation 2:
- Substitute:
[tex]\[
7b - 5\left(\frac{5}{4}b\right) = 9
\][/tex]
[tex]\[
7b - \frac{25}{4}b = 9
\][/tex]
Multiply throughout by 4 to eliminate fractions:
[tex]\[
28b - 25b = 36
\][/tex]
[tex]\[
3b = 36
\][/tex]
[tex]\[
b = 12
\][/tex]
Now that we have [tex]\( b = 12 \)[/tex], use Equation 1 to find [tex]\( r \)[/tex]:
- [tex]\( r = \frac{5}{4} \times 12 = 15 \)[/tex]
Thus, the original number of red counters is 15, and the original number of blue counters is 12.
