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). suppose a well is randomly selected from those in the county. let x: the number of impurities found in the well. find the probability distribution of x.

P(X=0) = 0.2, P(X=1) = 0.7, P(X=2) = 0.1

Step-by-step explanation:

Let's define the following events

A: the well has impurity A

B: the well has impurity B

: the complement of the event A

: the complement of the event B

we know that P(A) = 0.4, P(B) = 0.5 and P() = 0.2. Then

0.2 =  P() = (by De Morgan's laws)

= 1-P(), i.e., P() = 1-0.2 = 0.8

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

On the other hand, X is the number of impurities found in the well, so, X can take the values 0, 1 or 2.

X=2 is equivalent to "the well has both impurities"

P(X=2) = P() = 0.1

X=1 is equivalent to "the well has impurity A or impurity B but not both impurities"

P(X=1) = P() =  P()-P() (we used the fact that if C⊆D, then, P(D-C)=P(D)-P(C))

= 0.8-0.1=0.7

X=0 is equivalent to "the well has neither of the impurities"

P(X=0) = P() = 0.2

ok thats very interesting