Q:

people are infected with certain virus.. find $P(C | E^c)$

this is my problem and its solution:

The only thing I don't understand is how they came up with $P(E^c\cap C) = .074.$ I've tried adding numbers all over the place, multiplying them, I tried using the $P(C) = P(E \cap C) + P(E^c \cap C)$ rule (law of total probability) to find $P(E^c | C)$, but it came out negative. I thought $P(E^c \cap C)$ would be $P(E^c) \cdot P(C)$ but that comes out to be $.146$. 
Can someone tell me how they got $P(E^c \cap C) = .074?$

A:

Baye's Law states that for any event $A$ ad $B$, given $P(B)\neq0$, then 
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
In this context,
$$P(E^c|C)=\frac{P(E^c\cap C)}{P(C)}$$
$$P(E^c\cap C)=P(E^c|C)\cdot P(C)=0.37\cdot 0.2=0.074$$

