Why am I not allowed to use $P(A \land B) = P(A)P(B)$ formula? Question: 
A survey of $1000$ people determines that $80\%$ like walking and $60\%$ like biking, and
all like at least one of the two activities. What is the probability that a randomly chosen person in this survey likes biking but not walking?
What I did was use the formula:
$$P(A \land B) = P(A)P(B) = 0.8\cdot 0.6 = 0.48$$
$$P(B) - P(A  ∩ B) = P(A'  ∩ B) = 0.6 -0.48 = 0.12$$
but the answer should be $0.2$. 
In the answer key, $ P(A ∪ B) = P(A) + P(B) - P (A' ∩ B)$ is used instead. Why?
 A: $P(A\cap B) = P(A) P(B)$ only if $A$ and $B$ are independent. In your case, $A$ and $B$ are not independent, since you know that if a person doesn't like walking (i.e., you have information about $A$), then they must like biking (i.e., from that information, you can conclude something about $B$).
A: Here it is easier to draw a contingency table: From the given informations you get


*

*800 persons like walking

*600 person like biking

*1000 persons like at least one of walking and biking


This will end up in this table: 
\begin{array}{|c|c|c|}
\hline
&\mathrm{biking}&\text{not biking}\\
\hline
\text{walking}&400&400\\
\hline
\text{not walking}& 200&0\\
\hline
\end{array}
You see directly $P(\text{biking} \wedge \text{not walking})=\frac{200}{1000}=0.2$. To see that walking and biking are not independent, you only have to check whether
$P(\text{biking}\wedge \text{walking})$ equals $P(\text{biking})P(\text{walking})$ or not. But it's obvious $P(\text{biking})P(\text{walking})=0.6\cdot 0.8=0.48\neq 0.2=P(\text{biking}\wedge \text{walking})$.
