Chapter 2: Q. 2.35 (page 51)

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls.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(a) 3 red, 2 blue, and 2 green balls are withdrawn;
(b) at least 2 red balls are withdrawn;
(c) all withdrawn balls are the same color;
(d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.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

Expert verified

The respective probabilities are

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

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

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

Step by step solution

Given information

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

Part(a)

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

Part(b)

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

$P (R \geq 2) = 1 - P (R \leq 1) = 1 - P (R = 0) - P (R = 1) = 1 - \frac{(\begin{matrix} 34 \\ 7 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} - \frac{(\begin{matrix} 12 \\ 1 \end{matrix}) (\begin{matrix} 34 \\ 6 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} \Rightarrow P (R \geq 2) = \frac{363707}{608235}$

Part(c)

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

p = \frac{(\begin{matrix} 12 \\ 7 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} + \frac{(\begin{matrix} 16 \\ 7 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} + \frac{(\begin{matrix} 18 \\ 7 \end{matrix})}{(\begin{matrix} 46 \\ 7 \end{matrix})} p = \frac{5507}{6690585}

Part(d)

P(either exactly 3 red balls or exactly 3 blue balls are withdrawn) = P(3R or 3B)
$P (3 R or 3 B) = P (3 R) + P (3 B) P (3 R or 3 B) = \frac{(\begin{matrix} 12 \\ 3 \end{matrix})}{(\begin{matrix} 46 \\ 3 \end{matrix})} + \frac{(\begin{matrix} 16 \\ 3 \end{matrix})}{(\begin{matrix} 46 \\ 3 \end{matrix})} \Rightarrow P (3 R or 3 B) = \frac{13}{892078}$

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Part(a)

Part(b)

Part(c)

Part(d)

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