$P(B)=2/3, P(C)=3/4, P(A|C)=1/6, P(A|B \cap C)=1/12$, 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!
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.