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.

Answer 1

Answer:

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 )