There are some red counters and some blue counters in a bag. The ratio of red counters to blue counters is 4:1. Two counters are removed at random. The probability that both counters taken are red is \(\frac{22}{35}\). How many blue counters are in the bag?

Let's call the total number of counters in the bag as "n". Based on the ratio, we know that there are 4 red counters and 1 blue counter, so the total number of counters would be 4 + 1 = 5.

Let's assume there are 4x red counters and x blue counters in the bag. The total number of combinations of two counters from the bag would be (4x + x)C2, which is equal to (5x)C2 = (5x)(4x-1)/2.

The number of combinations where two red counters are selected would be 4xC2, which is equal to 4x(3x-1)/2. From the given information, the probability of taking two red counters is 22/35, so we can write an equation: 22/35 = 4xC2 / (5x)C2

Expanding both sides, we get: 22/35 = 4x(3x-1) / (5x)(4x-1)

Cross multiplying and simplifying, we get: 35(3x-1) = 22(5x)(4x-1)

Expanding both sides, we get: 105x - 35 = 110x^2 - 44x

Solving for x, we get: x = 5

So, there are 4x = 4 * 5 = 20 red counters and x = 5 blue counters in the bag.

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