Find the following probabilities using given ones I have to find the following probabilities in this exercise:
a) Find $P(A)$ when $P(A \cap B) = p_1$ and $P(A \cap \bar{B}) = p_2$
I think this is easy, but to make sure I am correct, I am posting the solution below, so please tell me if I am right or wrong:
$P(A) = P(A \cap B) + P(A \cap \bar{B}) = p_1 + p_2$
b) Find $P(\bar{A} \cap B)$ when $P(A)=p_1$ and $P(B)=p_2$
I did the following actions:
\begin{align*}
P(\bar{A} \cap B) &= P(B) - P(A \cap B) \\
P(\bar{A} \cap B) &= P(A \cup B) - P(A) \\
P(B) - P(A \cap B) &= P(A \cup B) - P(A) \\
P(B) - P(A \cap B) &= P(A) + P(B) - P(A \cap B) - P(A)
\end{align*}
Everything here gets simplified, so I can't find anything.
c) Find $P(A \vert B)$ and $P(B \vert A)$ when $P(A)=p_1$, $P(B)=p_2$ and $P(\bar{A} \cap \bar{B})=p_3$.
I have no idea about this one.
d) Prove that $P(B \vert A) \geq 1 - \frac{P(\bar{B})}{P(A)}$ and $P(B \vert A) \geq \frac{P(A) + P(B) - 1}{P(A)}$
I don't know how to solve this one too.
It would be awesome if you guys could help me. Thanks in advance!
 A: Basically what you need to solve these exercises is Bayes' rule
$$ P(A \vert B) = \frac{P(A \cap B)}{P(B)} $$

For c) start from Bayes' rule and note that
$$P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B)$$
and 
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
you can then write
\begin{align}
 P(A \cap B) &= P(A) + P(B) - P(A \cup B)\\
             &= P(A) + P(B) + P(\bar{A} \cap \bar{B}) - 1
\end{align}
and finally
$$ P(A \vert B) = \frac{P(A) + P(B) + P(\bar{A} \cap \bar{B}) - 1}{P(B)} $$
Use a similar reasoning for $P(B \vert A)$.

For d) 1) 
Start from Bayes ' rule
$$ P(B \vert A) = \frac{P( A \cap B)}{P(A)} $$
and notice that
$$ P(A \cap B) = P(A) + P(B) - P(A \cup B)$$
so that 
\begin{align}
P(B \vert A) &= \frac{P(A) + P(B) - P(A \cup B)}{P(A)} \\
             &= \frac{P(A) + (1-P(\bar{B})) - P(A \cup B)}{P(A)} \\
             &= \frac{P(A) - P(\bar{B}) - (P(A \cup B)-1)}{P(A)} \\
             &= {\color{blue}{1 - \frac{P(A \cup B)-1}{P(A)}}} - \frac{P(\bar{B})}{P(A)}
\end{align}
because $P(A \cup B) \leq 1$ and $P(A) \geq 0$ we have 
$$ \frac{P(A \cup B)-1}{P(A)} \leq 0 $$ 
hence
$$ {\color{blue}{1 - \frac{P(A \cup B)-1}{P(A)}}} {\color{red}{\geq 1}} $$
and going back to our formula for the conditional probability
\begin{align}
P(B \vert A) &= \left({\color{blue}{1 - \frac{P(A \cup B)-1}{P(A)}}}\right) - \frac{P(\bar{B})}{P(A)} \\
 &{\color{red}{\geq 1}} - \frac{P(\bar{B})}{P(A)} 
\end{align}
For d) 2)
Start from Bayes' rule
$$ P(B \vert A) = \frac{P( A \cap B)}{P(A)} $$
and notice that
$$ P(A \cap B) = P(A) + P(B) - P(A \cup B)$$
Now since $P(A \cup B) \leq 1$
$$ P(B \vert A) = \frac{ P(A) + P(B) - P(A \cup B) }{P(A)}  \geq  \frac{ P(A) + P(B) - 1 }{P(A)} $$
