osnomu3

2021-12-31

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.

Rita Miller

Expert

Step 1
We consider the four possible cases of the initial draw.
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.
${p}_{1}=\frac{\left(\begin{array}{c}9\\ 2\end{array}\right)\cdot 6}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$
Case 2: 2 used balls are drawn.
${p}_{2}=\frac{9\cdot \left(\begin{array}{c}6\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$
Case 3: 3 used balls are drawn.
${p}_{3}=\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$
Step 2
We evaluate the probabilities and outcomes of each case.
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.
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.
${p}_{0}\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+{p}_{1}\frac{\left(\begin{array}{c}7\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+{p}_{2}\frac{\left(\begin{array}{c}8\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+{p}_{3}\frac{\left(\begin{array}{c}9\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$
Step 4
$.1846\cdot .044+.4747\cdot .0769+.2967\cdot .1231+.044\cdot .1846=.0893$

Timothy Wolff

Expert

Probability to choose 3 not previously used balls from 15 when 9 are not previously used:
$\frac{9\cdot 8\cdot 7}{15\cdot 14\cdot 13}$
Conditionally on this, probability that these 3 balls were not played with because they would have been chosen in the first phase:
$\frac{12\cdot 11\cdot 10}{15\cdot 14\cdot 13}$
Thus, the desired probability is
$\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7}{{\left(15\cdot 14\cdot 13\right)}^{2}}\approx 0.893$.

Vasquez

Expert

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 $P\left(E1\right)=\sum _{n=0}^{3}\left(\begin{array}{c}9\\ n\end{array}\right)\cdot \frac{\left(\begin{array}{c}6\\ 3-n\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$
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 $P\left(E2\right)=\sum _{n=0}^{3}\frac{\left(\begin{array}{c}9-n\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$. 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 $P\left(E2\right)=\left(\frac{9-1c3}{15c3}\right)$ [9-1 because out of 9 unused balls, 1 ball is chosen]
Now total probability $=P\left(E1\right)\cdot P\left(E2\right)$
$=\sum _{n=0}^{3}\left(\begin{array}{c}9\\ n\end{array}\right)\cdot \frac{\left(\begin{array}{c}6\\ 3-n\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}\cdot \frac{\left(\begin{array}{c}9-n\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$=0.089

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