There are some blue and red counters in a bag in the ratio 1:3. Two counters are taken at random. The probability of both counters being blue is 1/20. How many counters were in the bag?

Let N be the number of blue counters. This implies that 3N counters are red, and there are 4N counters in total.Assuming that you don't reinsert the first counter, for the first pick, you have N blue counters over 4N total counters, so you'll pick a blue counter with probability[tex]\dfrac{N}{4N}=\dfrac{1}{4}[/tex]For the second pick, you're left with N-1 blue counters over 4N-1 counters, so you'll pick a blue counter with probability[tex]\dfrac{N-1}{4N-1}[/tex]The probability of picking two blue counters with two picks is the product of the two probabilities:[tex]\dfrac{1}{4}\cdot\dfrac{N-1}{4N-1}[/tex]And we want this to equal 1/20, so we have[tex]\dfrac{1}{4}\cdot\dfrac{N-1}{4N-1}=\dfrac{1}{20} \iff \dfrac{N-1}{4N-1}=\dfrac{1}{5} \iff 5(N-1)=4N-1[/tex]We can expand the left hand side and solve for N:[tex]5(N-1)=4N-1 \iff 5N-5=4N-1 \iff N = 4[/tex]So, there are 4 blue counters and 12 red counters, for a total of 16 counters in the bag.We can indeed verify that the probabilities of picking the two blue counters is[tex]\dfrac{4}{16}\cdot \dfrac{3}{15} = \dfrac{1}{4}\cdot\dfrac{1}{5}=\dfrac{1}{20}[/tex]