Certainly! Let’s solve this problem step-by-step.
Step 1: Define the variables
- Let [tex]\( n \)[/tex] be the total number of counters in the bag.
- There are 4 red counters.
- The number of blue counters is [tex]\( n - 4 \)[/tex].
Step 2: Calculate the probability of drawing two blue counters consecutively without replacement.
- The probability of drawing one blue counter first is given by:
[tex]\[ \frac{\text{number of blue counters}}{\text{total number of counters}} = \frac{n - 4}{n} \][/tex]
- Once the first blue counter is drawn, there are now [tex]\( n - 1 \)[/tex] counters left and [tex]\( n - 5 \)[/tex] blue counters remaining.
Thus, the probability of drawing another blue counter is:
[tex]\[ \frac{\text{remaining blue counters}}{\text{remaining total counters}} = \frac{n - 5}{n - 1} \][/tex]
Step 3: Combine these probabilities to find the probability of drawing two blue counters:
[tex]\[ \text{Probability of two blue counters} = \left( \frac{n - 4}{n} \right) \times \left( \frac{n - 5}{n - 1} \right) \][/tex]
Step 4: Set this probability equal to the given probability of [tex]\( \frac{1}{3} \)[/tex]:
[tex]\[ \left( \frac{n - 4}{n} \right) \times \left( \frac{n - 5}{n - 1} \right) = \frac{1}{3} \][/tex]
Step 5: Solve the equation:
- Multiply both sides by [tex]\( 3 \)[/tex] to clear the fraction:
[tex]\[ 3 \left( \frac{(n - 4)(n - 5)}{n(n - 1)} \right) = 1 \][/tex]
- Simplify the left-hand side:
[tex]\[ \frac{3(n - 4)(n - 5)}{n(n - 1)} = 1 \][/tex]
- Multiply both sides by [tex]\( n(n - 1) \)[/tex] to clear the denominator:
[tex]\[ 3(n - 4)(n - 5) = n(n - 1) \][/tex]
- Expand and simplify each side:
[tex]\[ 3(n^2 - 9n + 20) = n^2 - n \][/tex]
[tex]\[ 3n^2 - 27n + 60 = n^2 - n \][/tex]
Step 6: Move all terms to one side to set the equation to zero:
[tex]\[ 3n^2 - 27n + 60 - n^2 + n = 0 \][/tex]
[tex]\[ 2n^2 - 26n + 60 = 0 \][/tex]
Step 7: Divide by the greatest common divisor:
[tex]\[ n^2 - 13n + 30 = 0 \][/tex]
This completes the derivation.
Thus, the derived equation is:
[tex]\[ n^2 - 13n + 30 = 0 \][/tex]
This equation correctly shows the relationship required in the problem statement.
