# There are 15 tennis balls in a box, of which 9 have not previously bee

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.

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Rita Miller
Step 1
We consider the four possible cases of the initial draw.
Case 0: No used balls are drawn.
$p_{0}=\frac{(\begin{array}{c}9\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}$
Case 1: 1 used ball is drawn.
$p_{1}=\frac{(\begin{array}{c}9\\ 2\end{array}) \cdot 6}{(\begin{array}{c}15\\ 3\end{array})}$
Case 2: 2 used balls are drawn.
$p_{2}=\frac{9 \cdot (\begin{array}{c}6\\ 2\end{array})}{(\begin{array}{c}15\\ 3\end{array})}$
Case 3: 3 used balls are drawn.
$p_{3}=\frac{(\begin{array}{c}6\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}$
Step 2
We evaluate the probabilities and outcomes of each case.
Case 0: $$\displaystyle{p}_{{{0}}}={.1846}$$, 6 new balls, 9 used left over.
Case 1: $$\displaystyle{p}_{{{1}}}={.4747}$$, 7 new balls, 8 used.
Case 2: $$\displaystyle{p}_{{{2}}}={.2967}$$, 8 new balls, 7 used.
Case 3: $$\displaystyle{p}_{{{3}}}={.044}$$, 9 new balls, 6 used.
Step 3
We assess the probability of each case times the probability that no used balls turn up in the second draw, and add everything up.
$p_{0} \frac{(\begin{array}{c}6\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+p_{1} \frac{(\begin{array}{c}7\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+p_{2} \frac{(\begin{array}{c}8\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+p_{3} \frac{(\begin{array}{c}9\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}$
Step 4
$$\displaystyle{.1846}\cdot{.044}+{.4747}\cdot{.0769}+{.2967}\cdot{.1231}+{.044}\cdot{.1846}={.0893}$$
###### Not exactly what youâ€™re looking for?
Timothy Wolff
Probability to choose 3 not previously used balls from 15 when 9 are not previously used:
$$\displaystyle{\frac{{{9}\cdot{8}\cdot{7}}}{{{15}\cdot{14}\cdot{13}}}}$$
Conditionally on this, probability that these 3 balls were not played with because they would have been chosen in the first phase:
$$\displaystyle{\frac{{{12}\cdot{11}\cdot{10}}}{{{15}\cdot{14}\cdot{13}}}}$$
Thus, the desired probability is
$$\displaystyle{\frac{{{12}\cdot{11}\cdot{10}\cdot{9}\cdot{8}\cdot{7}}}{{{\left({15}\cdot{14}\cdot{13}\right)}^{{{2}}}}}}\approx{0.893}$$.
Vasquez

We have 15 balls out of which 9 balls have not been previously used, so the remaining 6 balls are already used.
Let E1 be the first event when the 3 balls are randomly chosen and played with.
Now when choosing 3 balls, the combination can be choosing all 3 unused, 2 unused and 1 used, 1 unused and 2 used & all 3 used.
So $$P(E1)=\sum_{n=0}^{3} (\begin{array}{c}9\\ n\end{array}) \cdot \frac{(\begin{array}{c}6\\ 3-n\end{array})}{(\begin{array}{c}15\\ 3\end{array})}$$
Here n is the no of unused balls. For eg:- when n is 1, we can choose 1 unused ball from 9 unused in (9c1) ways and remaining 2 from 6 used balls in (6c2) ways .
Now let E2 be event of picking 3 unused balls after E1.
Then $$P(E2)=\sum_{n=0}^{3} \frac{(\begin{array}{c}9-n\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}$$. Here n is the no of unused balls chosen after event E1. If suppose in E1, 1 unused and 2 used balls were chosen then $$P(E2)=(\frac{9-1c3}{15c3})$$ [9-1 because out of 9 unused balls, 1 ball is chosen]
Now total probability $$=P(E1) \cdot P(E2)$$
$$=\sum_{n=0}^{3} (\begin{array}{c}9\\ n\end{array}) \cdot \frac{(\begin{array}{c}6\\ 3-n\end{array})}{(\begin{array}{c}15\\ 3\end{array})} \cdot \frac{(\begin{array}{c}9-n\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}$$=0.089