Statistics: Conditional Probability $P(A│B)=\frac25$  ,$P(B)=\frac14$, $P(A)=\frac13$. Find 


*

*$P(A\land B)$

*$P(B|A)$


Here is what I did:
Part 1.
$$P(A\land B) = P(A) \cdot P(B)\\
= \frac13\cdot\frac14=\frac{1}{12}$$
Part 2.
$$P(B|A) = \frac{P(B\land A)}{P(A)} = \frac{\frac1{12}}{\frac13}=\frac14$$
I'm not too sure if my answers are right and I will like them checked over. Thanks
 A: For part (a), you made a mistake. It is not true that $P(A\cap B)=P(A)\cdot P(B)$. This is only true if $A$ and $B$ are independent.
To solve the task, here is a hint:

What is the definition of $P(A|B)$?

For part (b), your solution is OK, but the result is wrong because you calculated $P(A\land B)$ wrong in the first step.
A: Just use the definition of conditional probability $$P(A|B) = \frac{P(A\cap B)}{P(B)}$$


*

*$P(A\cap B) = P(A|B)P(B) = \frac25 \cdot \frac14 = \frac{1}{10}$

*$P(B|A) = \frac{P(A\cap B)}{P(A)} = \frac{\frac{1}{10}}{\frac13} = \frac{3}{10}$

A: You can't assume $P(A\cap B)=P(A)P(B)$ unless $A$ and $B$ are independent which they aren't.  Instead use $p(A|B)=P(A\cap B)/P(B)$.  You are given two of those values so you just need to plug those in to solve for the one you don't know.
A: Start with the definition of $P(A \cap B)$:
$P(A \cap B) = P(A|B) \times P(B) = P(B|A) \times P(A)$
Since you know $P(A|B), P(B)$ and $P(A)$ you can easily deduce the rest.
As Gregory mentioned, you are incorrectly assuming that A and B are independent i.e. $P(A|B) = P(A)$ and $P(B|A) = P(B)$.
