Q. 3.29 聽 There are 15 tennis balls in ... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 3: Q. 3.29 (page 99)

There are 15 tennis balls in a box, of which 9 have not previously been used. Three of the balls are randomly chosen, played with, and then returned to the box. Later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used.

Short Answer

Expert verified

The likelihood is that none of these balls have ever been used is $.0893$

Step by step solution

Step1:Given data

Case $0$ : No used balls are drawn. $p_{0} = \frac{(\begin{matrix} 9 \\ 3 \end{matrix})}{(\begin{matrix} 15 \\ 3 \end{matrix})}$

Case $1 : 1$ used ball is drawn. $p_{1} = \frac{(\begin{matrix} 9 \\ 2 \end{matrix}) 路 6}{(\begin{matrix} 15 \\ 3 \end{matrix})}$

Case $2 : 2$ used balls are drawn. $p_{2} = \frac{9 路 (\begin{matrix} 6 \\ 2 \end{matrix})}{(\begin{matrix} 15 \\ 3 \end{matrix})}$

Case $3 : 3$ used balls are drawn. $p_{3} = \frac{(\begin{matrix} 6 \\ 3 \end{matrix})}{(\begin{matrix} 15 \\ 3 \end{matrix})}$

Step2:Three of the balls are chosen at random.

Case $0 : p_{0} = . 1846, 6$ new balls, $9$ used left over.

Case $1 : p_{1} = . 4747, 7$ new balls, $8$ used.

Case $2 : p_{2} = . 2967, 8$ new balls, $7$ used.

Case $3 : p 3 = . 044, 9$ new balls, $6$ used.

Another 3 ball is drawn at random from the box.

$p_{0} \frac{(\begin{matrix} 6 \\ 3 \end{matrix})}{(\begin{matrix} 15 \\ 3 \end{matrix})} + p_{1} \frac{(\begin{array}{l} 7 \\ 3 \end{array})}{(\begin{matrix} 15 \\ 3 \end{matrix})} + p_{2} \frac{(\begin{array}{l} 8 \\ 3 \end{array})}{(\begin{matrix} 15 \\ 3 \end{matrix})} + p_{3} \frac{(\begin{array}{l} 9 \\ 3 \end{array})}{(\begin{matrix} 15 \\ 3 \end{matrix})}$