Conditional probability problem help P(A)=0.1, P(A and B)=0.05, P[(not A) and (not B)]=0.55. Ask P(A given B)
I don't know which formula or how do I start it, can anyone help?
 A: $$P(A'\cap B') = 0.55 = P((A\cup B)')\implies P(A\cup B)=0.45$$
$$P(A)+P(B)=P(A\cup B)+P(A\cap B) \implies P(B)=P(A\cup B)+P(A\cap B)-P(A)=0.45+0.05-0.1=0.4$$
So finally $$P(A|B)={P(A\cap B)\over P(B)} = 0.05/0.4=0.125$$
A: Hints: $P(\overline A \cap B) = P( \overline A)-P(\overline A \cap \overline B)$
$P(\overline A)=1-0.1=0.9$, $P(\overline A \cap \overline B)=0.55$
And $P(B)=P(\overline A \cap B) + P( A \cap B)$
Finally $P(A | B)=\frac{P( A \cap B)}{P(B)}$
A: Recall the rule for conditional probability: $$P(A|B) = \frac{P(A \land B)}{P(B)}$$
Now recall the rule of inverse probability: $P(A^C) =  1-P(A)$. So: $$P(((\lnot A) \land (\lnot B))^C) = 1-0.55$$
You should know from elementary logic/set theory that $((\lnot A) \land (\lnot B))^C = A \lor B$. By the above, $A \lor B = 1-0.55 = 0.45$.
Another useful fact: $P(A \lor B) = P(A) + P(B) - P(A \land B)$. From this we get
$$P(B) = P(A \lor B) + P(A \land B) - P(A) = 0.45 + 0.05 - 0.1 = 0.4$$
So by the rule of conditional probability we have
$$P(A|B) = \frac{P(A \land B)}{P(B)} = \frac{0.05}{0.4} = \frac 1 8$$
