A question about calculating conditional probability $P(B)=2/3, P(C)=3/4, P(A|C)=1/6, P(A|B \cap C)=1/12$, finds:
(a)$ P(C|B)$?
(b)$ P(A \cap C|B)$
(c) $P(\overline{B}|A \cap C)$
I only know that $P(A|C)=1/6 = P(A\cap C)/P(C)$ so $P(A\cap C) = 1/8$.
$P(A|B \cap C)=1/12 = P(A \cap B \cap C)/P(B \cap C) = 1/12,$ but we do not know $P(A \cap B \cap C)$ and we don't know $P(B \cap C).$
I got stuck here. How to approach this question?
Thank you very much!
 A: You've not given enough information to calculate any of the three quantities in (a), (b) or (c).  Let $\ \alpha,\beta,\gamma\ $ be any real numbers satisfying
\begin{align}
\frac{5}{144}&<\alpha<\frac{5}{88}\\
0&<\beta<\frac{2}{3}-12\alpha\\
0&<\gamma<12\alpha-\frac{5}{12}
\end{align}
and
\begin{align}
p_{000}&=\alpha\\
p_{001}&=\beta\\
p_{010}&=\frac{1}{8}-\alpha\\
p_{011}&=\gamma\\
p_{100}&=11\alpha\\
p_{101}&=\frac{2}{3}-12\alpha-\beta\\
p_{110}&=\frac{5}{8}-11\alpha\\
p_{111}&=12\alpha-\gamma-\frac{5}{12}\ .
\end{align}
Then $\ p_\omega\ $ is a probability mass function on the set
$$
\Omega=\{000,001,010,011,100,101,110,111\}\ .
$$
Now let
\begin{align}
A&=\{000,001,010,011\}\\
B&=\{000,001,100,101\}\ \ \text{and}\\
C&=\{000,010,100,110\}\ .
\end{align}
Then
\begin{align}
P(B)&=p_{000}+p_{001}+p_{100}+p_{101}\\
&=\frac{2}{3}\\
P(C)&=P_{000}+p_{010}+p_{100}+p_{110}\\
&=\frac{3}{4}\\
P(A\cap C)&=p_{000}+p_{010}\\
&=\frac{1}{8}\\
P(A\cap B\cap C)&=p_{000}\\
&=\alpha\\
P(B\cap C)&=p_{000}+p_{100}\\
          &=12\alpha\\
P(A\,|\,B\cap C)&=\frac{P(A\cap B\cap C)}{P(B\cap C)}\\
&=\frac{1}{12}\ .
\end{align}
Thus, these events in this probability space satisfy all the identities given, but
\begin{align}
P(B\,|\,C)&=\frac{P(B\cap C)}{P(C)}\\
&=16\alpha\\
P(A\cap B\,|\,C)&=\frac{P(A\cap B\cap C)}{P(C)}\\
&=\frac{4\alpha}{3}\\
P(\overline{B}\cap A\cap C)&=p_{010}\\
&=\frac{1}{8}-\alpha\\
P(\overline{B}\,|
 A\cap C)&=\frac{P(\overline{B}\cap A\cap C)}{P(A\cap C)}\\
&=1-8\alpha\ .
\end{align}
Since $\ \alpha\ $ can assume any value in the interval $\ \left(\frac{5}{144},\frac{5}{88}\right)\ $, these quantities are clearly not determined by those you've been given.
