Question

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?

Short answer

$P\left(R\right)=\frac{24722}{28275}$and

$P\left(B\right|\overline{A})=\frac{675}{2261}$

Step-by-step solution

Given information

Urn contains 8 Red, 10 Green and 12 Blue.

6 balls are selected at random.

Part(a)

Let Abe the event that atleast one red ball is chosen. Thus,

$$\begin{gathered} P\left(A\right)=1-P\left(\overline{A}\right)=1-\frac{\left(\begin{array}{c}22\\ 6\end{array}\right)}{\left(\begin{array}{c}30\\ 6\end{array}\right)} \\ \Rightarrow P\left(A\right)=\frac{24722}{28275} \end{gathered}$$

Part(b)

Let Bbe the event that exactly 2 green balls are chosen. Therefore,

$P\left(B\right|\overline{A})=\frac{\left(\begin{array}{c}10\\ 2\end{array}\right)\left(\begin{array}{c}12\\ 4\end{array}\right)}{\left(\begin{array}{c}22\\ 6\end{array}\right)}=\frac{675}{2261}$