Answer:
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
[tex]A^{c}[/tex]: the complement of the event A
[tex]B^{c}[/tex]: the complement of the event B
we know that P(A) = 0.4, P(B) = 0.5 and P([tex]A^{c}\cap B^{c}[/tex]) = 0.2. Then
0.2 = Â P([tex]A^{c}\cap B^{c}[/tex]) = [tex]P[(A\cup B)^{c}][/tex] (by De Morgan's laws)
= 1-P([tex]A\cup B[/tex]), i.e., P([tex]A\cup B[/tex]) = 1-0.2 = 0.8
and P([tex]A\cap B[/tex]) = P(A)+P(B)-P([tex]A\cup B[/tex]) = 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([tex]A\cap B[/tex]) = 0.1
X=1 is equivalent to "the well has impurity A or impurity B but not both impurities"
P(X=1) = P([tex](A\cup B)-(A\cap B)[/tex]) =  P([tex]A\cup B[/tex])-P([tex]A\cap B^[/tex]) (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([tex]A^{c}\cap B^{c}[/tex]) = 0.2
