High School

We appreciate your visit to There are only tex r tex red counters and tex b tex blue 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]$b$[/tex] blue counters in a bag.

A counter is taken at random from the bag. The probability that the counter is blue is [tex]$\frac{4}{9}$[/tex]. The counter is put back in the bag.

6 more red counters and 3 more blue counters are added to the bag.

A counter is taken at random from the bag. The probability that the counter is blue is [tex]$\frac{5}{12}$[/tex].

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

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.

Thanks for taking the time to read There are only tex r tex red counters and tex b tex blue 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