Using $P(A'|B')$ to find $P(A\cup B)$

Question

We have $P(B) = \frac 3 5$ and $P(A'|B') = \frac 1 3$.

I started with $P(A\cap B)' = P(B)' \cdot P(A'|B') = \frac 2 5 \cdot \frac 1 3 = \frac 2 {15}$.

Then converted $P(A\cap B)'$ to $P(A'\cup B')$.

Is $P(A'\cup B') = P(A\cup B)'$, meaning we get $P(A\cup B) = 1 - \frac 2 {15} = \frac {13} {15}$?

Answer 1

You have made two mistakes, but got the right answer! $P(A'\cap B')=P(A'|B') P(B')=\frac 1 3 (1-\frac 3 5)= \frac 2 {15}$. $P(A\cup B)=1-P(A' \cap B')=1-\frac 2 {15}=\frac {13} {15}$.

Answer 2

$\frac{1}{3}=P(A^c/B^c)=\frac{P(A^c \cap B^c)}{P(B^c)}=\frac{P((A \cup B)^c)}{1-P(B)}=\frac{1-P(A \cup B)}{1-\frac{3}{5}}$ $$\implies \frac{1×2}{3×5}=1-P(A \cup B)$$ $$\implies P(A \cup B)=\frac{13}{15}$$