An urn contains $30$ white and $15$ black balls. If $10$ balls are drawn without replacement, find the probability that the first $2$ balls are white, given that the sample contains exactly $6$ white balls
Let, $E_1=\{\text{sample has 6 white balls}\}$, $E_2=\{\text{first two balls white}\}$
I have to find $\displaystyle\Pr(E_2|E_1)=\frac{\Pr(E_1\cap E_2)}{\Pr(E_1)}$
$\displaystyle\Pr(E_1)=\frac{\binom{30}{6}\binom{15}{4}}{\binom{45}{10}}$ and
$\displaystyle\Pr(E_1\cap E_2)=\frac{\frac{30}{45}\frac{29}{44}\binom{28}{4}\binom{15}{4}}{\binom{45}{10}}$
The result should be $\frac13$, so I must have made a mistake, can you help please, Thanks.
Given that you have 6 white and 4 black balls, how many ways are there to line them up with place 2 white balls in the first two places?
$${8\choose 4}=70$$
Given that you have 6 white and 4 black balls, how many ways are there to line them up in total?
$${10\choose 6}=210$$
Divide the former by the latter and you have: $$P(\text{2 white balls drawn first} \mid 6\text{ white and $4$ black drawn}) = \tfrac{1}{3}$$
Addendum:
For your method, you have correctly counted: $$\Pr(E_1) = \frac{{30\choose 6}{15\choose 4}}{{45\choose 10}} = \dfrac{6927375}{27266558}\quad \checkmark$$
However, for the next part you should have counted thus: $$\Pr(E_1\cap E_2) = \dfrac{{30\choose 2}{28\choose 4}{15\choose 4}}{{45\choose 2}{43\choose 8}} = \dfrac{2309125}{27266558} = \frac 1 3 \Pr(E_1)$$
Count in the numerator the ways to choose 2 of 30 white balls for the first two places, and then 4 of the remaining 28 white balls and 4 of 15 black balls.
Count in the denominator the ways to choose 2 of 45 ball for the first two places, and then 8 of the remaining 43 balls.
What you count for the denominator must always correspond to the general case of what you count for the numerator.