Given, $P(A)=0.4, P(B)=0.5,P(A\mid B)=0.3 $.
Need to find $$P(A\cap B), \,P(B\mid A),\,P(A\cup B).$$
So far I did $$P(A\cap B) = P(A\mid B) P(B) = 0.15.$$
And I have the formula $$P(B\mid A)= \frac{P(A\mid B)P(B)}{P(A\mid B)P(B) + P(A\mid\bar B)P(\bar B)}$$
How do I find $P(A|\bar B)$? And what is the formula for $P(A\cup B)$? Please help.
Here is an approach.
From $$ P_{B}(A)=\frac{P(A\cap B)}{P(B)} $$ you deduce that $$ P(A\cap B)=P(B)\times P_{B}(A) $$ giving $$ P(A\cap B)=0.5\times0.3=\color{blue}{0.15} $$ From $$ P_{A}(B)=\frac{P(A\cap B)}{P(A)} $$ you deduce that $$ P_{A}(B)=\frac{0.15}{0.4}=\color{blue}{0.375} $$ On the other hand, we have $$ P(A\cup B)=P(A)+P(B)-P(A\cap B) $$ giving $$ P(A\cup B)=0.4+0.5-0.15=\color{blue}{0.75}. $$ You may notice that $$ P_{\overline{B}}(A)=\frac{P(A\cap \overline{B})}{P(\overline{B})}=\frac{P(A)-P(A\cap B)}{1-P(B)}=\frac{0.4-0.15}{0.85}=\color{blue}{\frac{5}{17}}\approx \color{blue}{0.2941}. $$