Set Questions? Need someone to help show I tried doing these and I have the answers, but I can only do A. Can someone show me how to do the rest? I have the answers but don't know how to get to them..

 A: *

*$B\cup C=C$, so $\mathbb{P} B\cup C = \mathbb{P} C = 0.7$.

*$A\cap C = A$. By the properties of conditional expectation, $$\mathbb{P} A = \mathbb{P}[A\mid B]\cdot \mathbb{P}B = \mathbb{P}[A\mid B]\cdot \mathbb{P}[B\mid C]\cdot \mathbb{P} C = 0.5\cdot 0.6\cdot 0.7 = 0.21$$

*Using the definition of conditional probabilities,
$$\mathbb{P}[B\mid \bar{A}] = \frac{\mathbb{P}( B\cap \bar{A})}{ \mathbb{P}\bar{A} } = \frac{ \mathbb{P}( B\setminus A)}{1-\mathbb{P}A} = \frac{\mathbb{P}B - \mathbb{P} A) }{ 1-\mathbb{P}A}$$ (the last one as $A\subseteq B$), which becomes, as $\mathbb{P} B =\mathbb{P}[B\mid C]\cdot \mathbb{P} C = 0.42$,
$$
\mathbb{P}[B\mid \bar{A}] = \frac{0.42-0.21}{1-0.21} = \frac{21}{79}
$$

*Similarly, 
$$\mathbb{P}[\bar{B}\mid A] = \frac{\mathbb{P} \bar{B}\cap A}{ \mathbb{P}A }$$
but $\bar{B}\cap A=\emptyset$, since $A\subseteq B$ and $B\cap \bar{B}=\emptyset$. So $$\mathbb{P}[\bar{B}\mid A] = \frac{\mathbb{P} \emptyset}{ 0.21 } = \frac{0}{0.21} = 0.$$

*Finally, since $A\subseteq B \subseteq C$, we have that $A\cap B \cap C = A$. Therefore, the ugly-looking quantity $\mathbb{P} \overline{A\cap B \cap C}$ simplifies a lot:
$$
\mathbb{P} \overline{A\cap B \cap C} = \mathbb{P} \bar{A} = 1-\mathbb{P} A = 1-0.21 = 0.79.
$$

