Calculate conditional probability; throwing the cube The symmetrical cube was threw $30$ times. 
Calculate probability situation, when in first $20$ throws came out $4$ times number $3$, if in $30$ throws number $3$ came out $7$ times.
 A: How many ways can the result $(3)$ occur $7$ times in $30$ throws?
$$\binom{30}{7}$$
How many ways can the result $(3)$ occur $4$ times in the first $20$ throws?
$$\binom{20}{4}$$
How many ways can the result $(3)$ occur $3$ times in the remaining $10$ throws?
$$\binom{10}{3}$$
The first question defines your sample space. The other two define your event. The final solution is:
$$P=\frac{\binom{20}{4}\binom{10}{3}}{\binom{30}{7}}=\frac{323}{1131}\approx0.285588$$
A: So in this case we are dealing with conditional probability, I think.
Let's calculate  $\mathbf{B}$
$7=7+0+0 \to $ $3$ times, because 7 could be anywhere
$7=6+1+0\to$ $6$ times
$7=5+1+1\to$ $3$ times 
$7=5+2+0\to$ $6$ times
$7=4+3+0\to$ $6$ times
$7=4+2+1\to$ $6$ times
$7=3+2+2\to$ $3$ times
$7=3+1+3\to$ $3$ times
So 
$\mathbf{B}=36$
Let's calculate $\mathbf{A} \cap\ \mathbf{B}$
$4=3+1\to$ $2$ times
$4=4+0\to$ $2$ times
$4=2+2\to$ $1$ time
$\mathbf{A} \cap\ \mathbf{B}=5$
So
$ P(\mathbf{A} \setminus \mathbf{B})=\left(\frac{5}{36}\right) $
Is it correct answer? Thanks in advance
