If $A\subset B$ then $P(A|B)=1$ because the probability of happening of $A$,given that $B$ has already happened is $1,$because $A$ is a subset of B. If $A\subset B,$then $P(A\mid B)=1$ because the probability of happening of $A$,given that $B$ has already happened is $1,$because $A$ is a subset of $B.$
but $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}\le 1$ because $P(A)\le P(B)$ as $A\subset B$ 
Why is there contradiction?Is my first statement correct that $P(A\mid B)=1?$.If not why?I am confused.Please help.
 A: 
If $A\subset B,$then $P(A\mid B)=1$ because the probability of happening of $A$,given that $B$ has already happened is $1,$because $A$ is a subset of $B.$

False.
If I know $B$ happened, and $A$ is a subset of $B$, I don't yet know $A$ happened, since there may be parts of $B$ that are not in $A$, and they also might have happened. However, if I know $A$ happened, then I know $B$ happened, because whatever in $A$ is also in $B$. In other words, $P(B\mid A)=1$, not $P(A\mid B)$. 
As you correctly wrote, $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}$$ which may not be $1$. On the other hand, $$P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{P(A)}{P(A)}=1.$$

For example, when rolling a die, set $A$ to be "rolling a $2$" and $B$ to be "rolling an even number".
Then, $A\subseteq B$ and $$P(A\mid B)=\frac{1}{3}$$ because if I know I rolled an even number, then I know I rolled a $2,4$ or $6$ - so the probability that I rolled a $2$ is one out of three.
On the other hand, $P(B\mid A)=1$ because if I know I rolled a $2$, then I also know I rolled an even number.
