91Ó°ÊÓ

$8$red, $10$green, $12$blue balls.

$6$of them are randomly chosen.

$R$ - at least $1$ red ball is chosen.

$A$ - exactly $2$ green balls are chosen.

$R^c$- no red balls are drawn

From the combinatorics - the number of ways to pick a combination of $6$balls out of $10+12$green and blue balls is:

The boxed formula above yields:

localid="1649691889561" $P\left(R^c\right)=\frac{\left(\begin{array}{c}22\\ 6\end{array}\right)}{\left(\begin{array}{c}30\\ 6\end{array}\right)}≈0.73$

This formula is derived from axioms of probability:

$P\left(R\right)=1-P\left(R^c\right)$

localid="1649691904394" $P\left(R\right)=1-0.73=0.27$.

$8$ red, $10$ green, $12$ blue balls.

$6$ of them are randomly chosen.

If $R^c$is given, that reduces the outcome space, and modifies the boxed formula into:

$P\left(Aâˆ\pounds R^c\right)=\frac{\#\text{outcomes in}A\text{which are in}{R}^c}{\#\text{of all possible outcomes in}{R}^c}$

For $A$to occur, choose $2$from $10$green balls and the remaining $4$from $12$blue balls (here$R^c$is used). From combinatorics, the general principle of counting and combinations show that:

Therefore:

localid="1649691934379" $P\left(Aâˆ\pounds R^c\right)=\frac{\left(\begin{array}{c}10\\ 2\end{array}\right)·\left(\begin{array}{c}12\\ 4\end{array}\right)}{\left(\begin{array}{c}22\\ 6\end{array}\right)}≈4.0$