When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 20% of the wells had neither impurity, 40% had impurity A, and 50% 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.

Question

Mathematics

Rigoberto Weeks

Yesterday, 01:08

When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 20% of the wells had neither impurity, 40% had impurity A, and 50% 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.

+5

Answers (1)

Know the Answer?

Kenna · Answer 1

P (Y = 0) = 0.2

P (Y = 1) = 0.7

P (Y = 2) = 0.1

Step-by-step explanation:

If 20% of wells didnt have impurities it could be said as that the probability that it didnt have A and didnt have B was 0.2:

P (not A) ∩ P (not B) = 0.2

Also:

P (A) = 0.4

P (B) = 0.5

The formulas for the union of both impurities are:

P (A) ∪ P (B) = P (A) + P (B) - P (A) ∩ P (B)

P (A) ∪ P (B) = 1 - P (not A) ∩ P (not B)

Replacing and solving you get:

P (A) + P (B) - P (A) ∩ P (B) = 1 - P (not A) ∩ P (not B)

0.4 + 0.5 - P (A) ∩ P (B) = 1 - 0.2

P (A) ∩ P (B) = 0.1

So the probability that the wells had both impurities is 0.1

P (A) ∪ P (B) = 0.8

This counts as having one of the impurities or both, in order to only have one impurity you need to substract when you have both:

one impurity = P (A) ∪ P (B) - P (A) ∩ P (B) = 0.8 - 0.1 = 0.7

Now Y is the probabilty distribuion for the numbers of impuritis.

P (Y = 0) = 0.2 (0 impurities)

P (Y = 1) = 0.7 (only 1 impurity)

P (Y = 2) = 0.1 (2 impurities)