When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 10% of the wells had neither impurity, 80% had impurity A, and 30% had impurity B. (Obviously, some had both impurities.) If a well is randomly chosen from those in the county, find the probability distribution for Y, the number of impurities found in the well.

To find the probability distribution for Y, the number of impurities found in a well, we need to calculate the probabilities for Y = 0, Y = 1, and Y = 2. We can use the information given to find these probabilities.

1. Y = 0 (No impurities) It's given that 10% of the wells had neither impurity. Therefore, the probability that a well has no impurities, P(Y = 0), is 0.10.

2. Y = 1 (One impurity) We aren't given the exact numbers for wells with only one impurity directly, but we can find this by considering the wells with each impurity and subtracting those with both.

Let's assume P(A) is the probability of a well having impurity A, P(B) is the probability of having impurity B, and P(A and B) is the probability of a well having both impurities.

It's given that: P(A) = 80% or 0.80 P(B) = 30% or 0.30 P(A and B) = P(A) + P(B) - P(A or B); where P(A or B) is 100% except the wells with neither, which is 10%.

P(A or B) = 1 - P(neither) = 1 - 0.10 = 0.90

Substituting the values we have: P(A and B) = 0.80 + 0.30 - 0.90 = 0.20 or 20%

The probability of a well having exactly one impurity is P(A) + P(B) - 2*P(A and B): P(Y = 1) = 0.80 + 0.30 - 2*(0.20) = 1.10 - 0.40 = 0.70 or 70%

3. Y = 2 (Both impurities) We've already calculated that the probability of a well having both impurities is P(A and B) which is 20% or 0.20.

So, the probability distribution for Y, the number of impurities in a well, is: P(Y = 0) = 0.10 P(Y = 1) = 0.70 P(Y = 2) = 0.20