There are some red counters and some blue counters in a bag. The ratio of red counters to blue counters is 5:1. Two counters are removed at random.

The probability that both counters taken are red is $\frac{22}{35}$.

The given information states that the ratio of red counters to blue counters in the bag is 5:1. Let's assume that there are 5x red counters and x blue counters in the bag. The total number of counters in the bag would then be 6x.

When 2 counters are randomly selected, the probability of selecting a red counter on the first draw is (5x / 6x), and the probability of selecting a red counter on the second draw, given that the first counter was red, is ((5x - 1) / (6x - 1)).

To calculate the probability of both counters being red, we multiply these two probabilities together:

(5x / 6x) * ((5x - 1) / (6x - 1)) = (25x^2 - 5x) / (36x^2 - 7x)

Given that this probability is equal to 22/35, we can set up the following equation:

(25x^2 - 5x) / (36x^2 - 7x) = 22/35

By solving this equation, we can find the value of x and determine the probability of both counters being red.

Learn more about probability here: brainly.com/question/13604758

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There are some red counters and some blue counters in a bag. The ratio of red counters to blue counters is 5:1. Two counters are removed at random.The probability that both counters taken are red is \(\frac{22}{35}\).

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There are some red counters and some blue counters in a bag. The ratio of red counters to blue counters is 5:1. Two counters are removed at random.

The probability that both counters taken are red is \(\frac{22}{35}\).