Find the probability of intersection 
Find $P(B \cap E)$ if $3$ digits are chosen in random with replacement from $0, 1, 2..., 9$ and $B = \{\text{All 3 digits are chosen are unique}\}$ and $E = \{ \text{All digits have the same parity (either odd or even)}\}$

I got that $P(B) = 0.72$ and $P(E) = 0.25$
This is a dependent probability, so how can I proceed? 
 A: You calculated $P(E)$ and $P(B)$ correctly. But it's important to understand that you just can not multiply $P(E)$ and $P(B)$ to get the answer!
Suppose you had $N$ experiments. Among those $N$ experiments there are approximately $N * P(B)$ experiments when all chosen digits have same parity. Consider only those experiments now. In some of these experiments all the digits are unique, in others they are not. In what part of these experiments all the digits are unique? Let's denote it X. Then the total expected number of experiments where both "all digits have same parity" and "all digits are unique" hold would be $N * P(B) * X$.
$X \ne P(E)$. They have different meaning! $P(E)$ is the probability that 3 random digits are unique, $X$ is the probability that 3 randomly selected digits of the same parity are unique. They are different.
Now back to original problem. I would suggest the following approach. There are 1000 different possibilities to choose 3 digits: 000, 001, ... 999.
Probability of any of them is the same: 1/1000.
Now manually count all the outcomes satisfying both $B$ and $E$. At least start doing it:
246, 264, ... - 6 combinations of digits (2, 4, 6)
and there are 6 combination for digits (0, 2, 6) and 6 combinations for...
Well we can choose three unique even digits in $\binom{5}{3}$ ways, so there are $3!*\binom{5}{3} = 60$ possible outcomes with unique even digits. And 60 possible outcomes with unique odd digits. Total $120$ outcomes of $1000$. So $0.12$ is the answer.
A: For dependent probability it holds:
$P(B∩E) = P(B)*P_B(E)$
or
$P(B∩E) = P(E)*P_E(B)$, where $P_B(E)$ is the probability $E$ is true, under the condition that $B$ is (already) true and vice versa. I am not posting the solution, since you wanted to proceed yourself!
A: As you stated this is a dependent probability and can be solved using bayes rule as follows.
Let we define two further events such that $E_o=\text{all digits are odd}$ and $E_e=\text{all digits are even}$. Thus $P(E)=P(E_o\cup E_e)$ Since , $P(E)$ can be written as, $P(E)=P(E_o)+P(E_e)$
So the original question becomes,
$$
P(B \cap E)=P(B \cap (E_o\cup E_e))=P((B \cap E_o)\cup (B \cap E_e))=P(B \cap E_o)+(B \cap E_e)
$$
Since $B \cap E_o$ and $B \cap E_o$ are disjoint events.
$P(B \cap E_o)$ can be computed using the Bayes rule:
$$
P(B \cap E_o)=P(B | E_o)*P(E_o)
$$
Where $P(E_o)=\frac{5*5*5}{10*10*10}=1/8$ (probability that all digits are odd) and $P(B | E_o)=\frac{5*4*3}{5*5*5}=12/25$ (probability that all digits are unique given that all numbers are odd. This decreases our sample space from 0,...,9 to 1,3,5,7,9)). Thus $P(B \cap E_o)=P(B | E_o)*P(E_o)=(1/8)*(12/25)=3/50$.
I believe $P(B \cap E_e)$ is same as $P(B \cap E_o)$ since they are symmetric. Thus the answer should be $3/50+3/50=3/25$
