College

We appreciate your visit to 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. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

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.

Your final line must say, "... red and ... green counters."

Answer :

Let's solve the problem step-by-step.

1. Initial Situation:
- There are [tex]\( r \)[/tex] red counters and [tex]\( g \)[/tex] green counters in a bag.
- The probability of picking a green counter initially is given by:

[tex]\[
\frac{g}{r + g} = \frac{4}{9}
\][/tex]

2. After Adding Counters:
- 4 more red counters and 2 more green counters are added to the bag.
- The total number of counters becomes [tex]\( r + 4 \)[/tex] red and [tex]\( g + 2 \)[/tex] green.
- The probability of picking a green counter now is:

[tex]\[
\frac{g + 2}{(r + 4) + (g + 2)} = \frac{10}{23}
\][/tex]

Now, let's set up the equations based on the given probabilities:

### First Equation:

From the initial probability,

[tex]\[
\frac{g}{r + g} = \frac{4}{9}
\][/tex]

Cross-multiply to clear the fraction:

[tex]\[
9g = 4(r + g)
\][/tex]

Simplify this:

[tex]\[
9g = 4r + 4g
\][/tex]

Subtract [tex]\( 4g \)[/tex] from both sides:

[tex]\[
5g = 4r
\][/tex]

Thus,

[tex]\[
g = \frac{4}{5}r \quad \text{(Equation 1)}
\][/tex]

### Second Equation:

From the probability after adding counters,

[tex]\[
\frac{g + 2}{r + g + 6} = \frac{10}{23}
\][/tex]

Cross-multiply:

[tex]\[
23(g + 2) = 10(r + g + 6)
\][/tex]

Expand both sides:

[tex]\[
23g + 46 = 10r + 10g + 60
\][/tex]

Simplify by subtracting [tex]\( 10g + 60 \)[/tex] from both sides:

[tex]\[
13g + 46 = 10r + 60
\][/tex]

Subtract 46 from both sides:

[tex]\[
13g = 10r + 14
\][/tex]

Now substitute the expression for [tex]\( g \)[/tex] from Equation 1 into this equation:

- Substitute [tex]\( g = \frac{4}{5}r \)[/tex] into [tex]\( 13g = 10r + 14 \)[/tex]:

[tex]\[
13 \left(\frac{4}{5}r\right) = 10r + 14
\][/tex]

This simplifies to:

[tex]\[
\frac{52}{5}r = 10r + 14
\][/tex]

Multiply every term by 5 to eliminate the fraction:

[tex]\[
52r = 50r + 70
\][/tex]

Subtract [tex]\( 50r \)[/tex] from both sides:

[tex]\[
2r = 70
\][/tex]

Divide both sides by 2:

[tex]\[
r = 35
\][/tex]

### Calculate [tex]\( g \)[/tex]:

Using Equation 1,

[tex]\[
g = \frac{4}{5} \times 35 = 28
\][/tex]

So, originally, there were 35 red counters and 28 green counters. Therefore, the final line is:

35 red and 28 green counters.

Thanks for taking the time to read 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. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada