High School

There are only [tex]r[/tex] red counters and [tex]g[/tex] green counters in a bag. A counter is taken at random from the bag. The probability that the counter is green is [tex]\frac{4}{9}[/tex]. The counter is put back in the bag. Four more red counters and two more green counters are added to the bag. A counter is taken from the bag again. The probability that the counter is green is [tex]\frac{10}{23}[/tex].

Find the number of red counters and the number of green counters that were in the bag originally.

Answer :

Sure, let's solve this probability problem step by step.

1. Understanding the initial setup:
- Let [tex]\( r \)[/tex] be the number of red counters and [tex]\( g \)[/tex] be the number of green counters originally in the bag.

2. First Condition:
- We are given that the probability of picking a green counter initially is [tex]\(\frac{4}{9}\)[/tex].
- This gives us the equation:
[tex]\[
\frac{g}{r + g} = \frac{4}{9}
\][/tex]
- Solving for [tex]\( g \)[/tex] in terms of [tex]\( r \)[/tex], we get:
[tex]\[
9g = 4(r + g)
\][/tex]
[tex]\[
9g = 4r + 4g
\][/tex]
[tex]\[
5g = 4r
\][/tex]
[tex]\[
g = \frac{4}{5}r
\][/tex]

3. Second Condition:
- After adding 4 more red counters and 2 more green counters, the number of red counters becomes [tex]\( r + 4 \)[/tex] and the number of green counters becomes [tex]\( g + 2 \)[/tex].
- The new probability of picking a green counter is [tex]\(\frac{10}{23}\)[/tex].
- This gives us the equation:
[tex]\[
\frac{g + 2}{r + 4 + g + 2} = \frac{10}{23}
\][/tex]
- Solve for this:
[tex]\[
\frac{g + 2}{r + g + 6} = \frac{10}{23}
\][/tex]
[tex]\[
23(g + 2) = 10(r + g + 6)
\][/tex]
[tex]\[
23g + 46 = 10r + 10g + 60
\][/tex]
[tex]\[
13g = 10r + 14
\][/tex]
[tex]\[
g = \frac{10}{13}r + \frac{14}{13}
\][/tex]

4. Solve the equations:
- Now, we have two equations:
- [tex]\( g = \frac{4}{5}r \)[/tex]
- [tex]\( g = \frac{10}{13}r + \frac{14}{13} \)[/tex]

- Set them equal to each other to solve for [tex]\( r \)[/tex]:
[tex]\[
\frac{4}{5}r = \frac{10}{13}r + \frac{14}{13}
\][/tex]

- Solve for [tex]\( r \)[/tex]:
[tex]\[
\frac{4}{5}r - \frac{10}{13}r = \frac{14}{13}
\][/tex]

- Get a common denominator (65):
[tex]\[
\frac{52}{65}r - \frac{50}{65}r = \frac{14}{13}
\][/tex]
[tex]\[
\frac{2}{65}r = \frac{14}{13}
\][/tex]
[tex]\[
r = \frac{14}{13} \times \frac{65}{2}
\][/tex]
[tex]\[
r = 35
\][/tex]

5. Finding [tex]\( g \)[/tex]:
- Substitute [tex]\( r = 35 \)[/tex] into [tex]\( g = \frac{4}{5}r \)[/tex]:
[tex]\[
g = \frac{4}{5} \times 35
\][/tex]
[tex]\[
g = 28
\][/tex]

Thus, the number of red counters originally in the bag is 35, and the number of green counters is 28.