Six balls are to be randomly chosen from an urn containing $8$ red, $10$green, and $12$blue balls.

(a) What is the probability at least one red ball is chosen?

(b) Given that no red balls are chosen, what is the conditional probability that there are exactly $2$ green balls among the $6$ chosen?

$8$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" $P\left(R^c\right)=\frac{\left(\begin{array}{c}22\\ 6\end{array}\right)}{\left(\begin{array}{c}30\\ 6\end{array}\right)}\approx 0.73$

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:

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

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" $P\left(A\mid R^c\right)=\frac{\left(\begin{array}{c}10\\ 2\end{array}\right)·\left(\begin{array}{c}12\\ 4\end{array}\right)}{\left(\begin{array}{c}22\\ 6\end{array}\right)}\approx 4.0$