So there's this game that I'm analysing, in which out of $45$ numbered balls ( numbered from $1$ to $45$ ), I choose $8$ balls.
$6$ out of the $45$ balls are drawn in the end of round by the organiser, and who get's $6$ out of $8$ balls of his draws, matching with the winning $6$ drawn balls, he wins the whole round.
my question is what is the probability of drawing those $6$ balls( $6$ winning numbers ), knowing that I drew 8 balls. So as a total we'll have $6$ correct matching balls and $2$ wrong. Order doesn't matter here.
My approach was that we have $6\times5\times4\times3\times2$ ways of drawing those $6$ balls, and $39 C 2$ ways to draw the $2$ wrong balls, over the all possible ways of drawing the $8$ balls out of the $45$ balls.
so my answer is $\frac{6\times5\times4\times3\times2 (39 C 2)}{45 C 8}$
Am I correct doing so ?
Your attempt is close to correct. The problem is that you are looking at all orders of choosing the six "correct" balls. Instead, there is only one way to choose all six of the six "correct" balls:
$$\dfrac{({_6C_6})({_{39}C_2})}{ {_{45}C_8} }$$