Q. 3.29 - A First Course in Probability

Question

  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.

Step-by-Step Solution

Answer

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

Step1:Given data

Case $0$: No used balls are drawn. $p_0=\frac{\left(\begin{array}{c}9\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

Case $1: 1$used ball is drawn. $p_1=\frac{\left(\begin{array}{c}9\\ 2\end{array}\right)·6}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

Case$2: 2$ used balls are drawn.$p_2=\frac{9·\left(\begin{array}{c}6\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

Case$3: 3$ used balls are drawn. $p_3=\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

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 $3: p3=.044,9$ new balls, $6$ used.

Step3: Another 3 ball is drawn at random from the box.

$p_0\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+p_1\frac{\left(\begin{array}{l}7\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+p_2\frac{\left(\begin{array}{l}8\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+p_3\frac{\left(\begin{array}{l}9\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

We multiply each case's probability by the probability that no used balls are found in the second draw, and then add it all up.

Step4:It's possible that none of these balls have ever been used.

$$\begin{gathered} .1846\times .044+.4747\times .0769+.2967\times .1231+.044\times .1846 \\ =.0893 \end{gathered}$$