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

Lets glance there at four likely choices of its first round.

$\text{Case 0:}$There are still no utilized balls displayed. $p_0=\frac{\left(\begin{array}{c}9\\ 3\end{array}\right)}{\left(\begin{array}{c}155\\ 3\end{array}\right)}$

$\text{Case}1:$There are still one utilized balls displayed. localid="1649414442532" $p_1=\frac{\left(\begin{array}{c}9\\ 2\end{array}\right)\times 6}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

$\text{Case 2:}$There are still two utilized balls displayed. localid="1649414452958" $p_2=\frac{9\times \left(\begin{array}{c}6\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

$\text{Case 3:}$There are still three utilized balls displayed.$p_3=\frac{\left(\begin{array}{l}6\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

Every trial's odds and possibilities was appraised.

$\text{Case 0:}{p}_0=.1846$,$6$new balls, $9$used.

$\text{Case}\text{:1}{p}_1=.4747,7$new balls,$8$used.

$\text{Case 2:}{p}_2=.2967,8$new balls, $7$used.

$\text{Case}3:p_3=.044,9$new balls,$6$used.

Simply raise the percentages of every instance by an occasion which no spent balls also willbe detected as in second draw, we add this together.

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

localid="1649414525255" $.1846\times .044+.4747\times .0769+.2967\times .1231+.044\times .1846=.0893$