College

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 put in the bag. A counter is taken from the bag. 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 :

To solve this problem, we need to find the original number of red counters ([tex]\(r\)[/tex]) and green counters ([tex]\(g\)[/tex]) in the bag based on the given probabilities.

1. Initial Probability Condition:
- We know the probability of picking a green counter initially is [tex]\(\frac{4}{9}\)[/tex].
- This can be expressed as:
[tex]\[
\frac{g}{r + g} = \frac{4}{9}
\][/tex]

2. Probability After Adding Counters:
- 4 more red counters and 2 more green counters are added to the bag.
- This changes the total number of red counters to [tex]\(r + 4\)[/tex] and green counters to [tex]\(g + 2\)[/tex].
- The total number of counters becomes [tex]\((r + 4) + (g + 2) = r + g + 6\)[/tex].
- The probability of selecting a green counter now is given as [tex]\(\frac{10}{23}\)[/tex], so:
[tex]\[
\frac{g + 2}{r + g + 6} = \frac{10}{23}
\][/tex]

3. Solving the Equations:
- We now have two equations:
1. [tex]\( \frac{g}{r + g} = \frac{4}{9} \)[/tex]
2. [tex]\( \frac{g + 2}{r + g + 6} = \frac{10}{23} \)[/tex]

- Solving these equations simultaneously will allow us to find the values of [tex]\(r\)[/tex] and [tex]\(g\)[/tex].

From these steps, the solution can be found:

- The original number of red counters ([tex]\(r\)[/tex]) is 35.
- The original number of green counters ([tex]\(g\)[/tex]) is 28.

These results match the conditions described in the problem, ensuring the probabilities align with the changes made to the bag.