1. **Problem statement:** There are red and blue counters in a bag with a ratio of red to blue counters as 3:1. Two counters are removed at random, and the probability that both are blue is $\frac{1}{20}$. We need to find the total number of counters in the bag before removal. 2. **Define variables:** Let the number of blue counters be $b$ and red counters be $r$. Given the ratio $r:b = 3:1$, we can write $r = 3k$ and $b = k$ for some positive integer $k$. 3. **Total counters:** Total counters $T = r + b = 3k + k = 4k$. 4. **Probability formula:** The probability of drawing two blue counters without replacement is $$ P = \frac{b}{T} \times \frac{b-1}{T-1} = \frac{1}{20} $$ 5. **Substitute values:** Substitute $b = k$ and $T = 4k$: $$ \frac{k}{4k} \times \frac{k-1}{4k-1} = \frac{1}{20} $$ 6. **Simplify:** $$ \frac{1}{4} \times \frac{k-1}{4k-1} = \frac{1}{20} $$ Multiply both sides by 20: $$ 20 \times \frac{1}{4} \times \frac{k-1}{4k-1} = 1 $$ Simplify: $$ 5 \times \frac{k-1}{4k-1} = 1 $$ 7. **Solve for $k$:** $$ 5(k-1) = 4k - 1 $$ $$ 5k - 5 = 4k - 1 $$ $$ 5k - 4k = -1 + 5 $$ $$ k = 4 $$ 8. **Find total counters:** $$ T = 4k = 4 \times 4 = 16 $$ **Answer:** There were 16 counters in the bag before any were removed.