Probability distribution

Question

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.

 

 

I dont understand how to get one on the answers. It says in my solutions booklet that the probability of getting both is: 

P( A and B ) = P(A) + (PB) - P( A union B) = 0.5 + 0.4- 0.8 = 0.1

 

I dont understand how you get this formula and I dont understand where you get P( A union B)= 0.8

Answer 1

The addition rule for probability goes as follows: P(A∪B) = P(A) + P(B) - P(A∩B). You can rearrange to get the formula used in your problem:  P(A∩B) = P(A) + P(B) - P(A∪B).

 

P(A∪B) is the probability that the well has one or both of the impurities. This would be one minus the probability that the well has neither impurity, 20%.

So, P(A∪B) = 1 - 0.20 = 0.80