Answer
$P(A|B) = \frac{1}{5} = 0.2$
Work Step by Step
Given
$P(A \cap B^c) = 0.3$,
$P((A \cup B)^c) = 0.2$
and $P(A \cap B) = 0.1$,
To find $P(A|B)$ we will use the formula:
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
Since $P(A\cap B)$ is given, we need to determine $P(B)$.
$P((A \cup B) = 1-P((A \cup B)^c)$ $= 1-0.2 = 0.8$
We know that
$P(A)=P(A\cap B)+P(A\cap B^c)$
$P(A)=P(A)+P(B)-P(A\cup B)+P(A\cap B^c)$
$P(B)=P(A\cup B)-P(A\cap B^c)=0.8-0.3=0.5$
Now we calculate $P(A|B)$:
$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.5} = \frac{1}{5} = 0.2$
