Events A and B are mutually exclusive. Suppose event A occurs with probability 0.39 and event B occurs with probability 0.32. If A does not occur, what is the probability that B does not occur?
I tried like you said (B|A¯)=P(B∩A¯)/P(A¯)=(0.32*0.61)/0.61= 0.119072=0.12 (Correct to 2 decimal places)
Is it Right?
On
$\mathbb P (\bar B|\bar A) = \frac{\mathbb P (\bar A \cap \bar B)}{\mathbb P (\bar A)} = \frac{\mathbb P (\overline{ A \cup B})}{\mathbb P (\bar A)} =\frac{1-\mathbb P ( A \cup B)}{1-\mathbb P (A)} = \frac{1-(\mathbb P ( A ) + \mathbb P (B))}{1-\mathbb P (A)} = 1 - \frac{\mathbb P (B)}{1-\mathbb P ( A )} = 1 - \frac{0.32}{0.61} = 0.48$
I think all this stuff is easier by diagrams. You're told this, thanks to $A$ and $B$ being m.e.
Since you are told (conditional prob.) that $A$ did not occur, your entire universe/denominator is out of $0.61$ instead of $1$. Out of this universe, you need to know how $B$ fails to occur. This is just the $0.29$, NOT including the $0.39$ because you know that $A$ did NOT happen. Hence the answer is $$ \frac{0.29}{0.61} \approx 0.4754. $$ In short, draw the diagram, then take your scissors and cut out the $A$ ring. Now proceed as usual. All you're using is $P(C \mid D) = \frac{P(C \cap D)}{P(D)}$ by thinking in terms of the diagram. I find this so much more intuitive than chasing formulas.