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 )
