P(B)=0.5, P(A or B)=0.3, and P(A and B)=0.6, find P(A)
How would you do this?
The answer is {P(A)= 0.4} but how in the world do you get that? How do you do it?
Thanks!
The problem contains inconsistent data.
For all events $\displaystyle X~\&~Y$ this is true: $\displaystyle X \subseteq Y\; \Rightarrow \;P(X) \leqslant P(Y)$.
But look at the given: $\displaystyle B \subseteq A \cup B\;\text{but} \;P(B) > P(A \cup B).$
So is impossible.
I suspect the question has been copied correctly but the professor who set the question has not thought past the following equation:
$\displaystyle \Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$
$\displaystyle \Rightarrow 0.3 = 0.5 + \Pr(B) - 0.6$
$\displaystyle \Rightarrow \Pr(B) = 0.4$.
@OP: Ask your professor how comfortable s/he is that it follows from the given data that $\displaystyle \Pr(B \cap A') = {\color{red}-} 0.1$.