Question

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that

(a) 3 red, 2 blue, and 2 green balls are withdrawn;

(b) at least 2 red balls are withdrawn;

(c) all withdrawn balls are the same color;

(d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Short answer

The respective probabilities are

(a)$P(3R+2B+2G)=\frac{3060}{40549}$

(b) $P(R\ge 2)=\frac{363707}{608235}$

(c) $p=\frac{5507}{6690585}$

(d)$\mathrm{P}(3\mathrm{R}\mathrm{or}3\mathrm{B})=\frac{13}{892078}$

Step-by-step solution

Given information

An urn that contains 12 red(R), 16 blue(B), and 18 green(G) balls.

Part(a)

$P(3R+2B+2G)=\frac{\left(\begin{array}{c}12\\ 3\end{array}\right)\left(\begin{array}{c}16\\ 2\end{array}\right)\left(\begin{array}{c}18\\ 2\end{array}\right)}{\left(\begin{array}{c}46\\ 7\end{array}\right)}=\frac{3060}{40549}$

Part(b)

P(atleast 2 red balls are drawn)$=P(R\ge 2)$

$$\begin{gathered} P(R\ge 2)=1-P(R\le 1) \\ =1-P(R=0)-P(R=1) \\ =1-\frac{\left(\begin{array}{c}34\\ 7\end{array}\right)}{\left(\begin{array}{c}46\\ 7\end{array}\right)}-\frac{\left(\begin{array}{c}12\\ 1\end{array}\right)\left(\begin{array}{c}34\\ 6\end{array}\right)}{\left(\begin{array}{c}46\\ 7\end{array}\right)} \\ \Rightarrow P(R\ge 2)=\frac{363707}{608235} \end{gathered}$$

Part(c)

P(all balls are of same color) = P(all R) + P(all B) + P(all G) = $p$(say)

$$\begin{gathered} p=\frac{\left(\begin{array}{c}12\\ 7\end{array}\right)}{\left(\begin{array}{c}46\\ 7\end{array}\right)}+\frac{\left(\begin{array}{c}16\\ 7\end{array}\right)}{\left(\begin{array}{c}46\\ 7\end{array}\right)}+\frac{\left(\begin{array}{c}18\\ 7\end{array}\right)}{\left(\begin{array}{c}46\\ 7\end{array}\right)} \\ p=\frac{5507}{6690585} \end{gathered}$$

Part(d)

P(either exactly 3 red balls or exactly 3 blue balls are withdrawn) = P(3R or 3B)
$$\begin{gathered} \mathrm{P}(3\mathrm{R}\mathrm{or}3\mathrm{B})=\mathrm{P}\left(3\mathrm{R}\right)+\hspace{0.17em}\mathrm{P}\left(3\mathrm{B}\right) \\ \mathrm{P}(3\mathrm{R}\mathrm{or}3\mathrm{B})=\frac{\left(\begin{array}{c}12\\ 3\end{array}\right)}{\left(\begin{array}{c}46\\ 3\end{array}\right)}+\frac{\left(\begin{array}{c}16\\ 3\end{array}\right)}{\left(\begin{array}{c}46\\ 3\end{array}\right)} \\ \Rightarrow \mathrm{P}(3\mathrm{R}\mathrm{or}3\mathrm{B})=\frac{13}{892078} \end{gathered}$$