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

Beginner2022-01-01Added 28 answers

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:${p}_{0}=.1846$ , 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.

$${p}_{0}\frac{(\begin{array}{c}6\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+{p}_{1}\frac{(\begin{array}{c}7\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+{p}_{2}\frac{(\begin{array}{c}8\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}+{p}_{3}\frac{(\begin{array}{c}9\\ 3\end{array})}{(\begin{array}{c}15\\ 3\end{array})}$$

Step 4

$.1846\cdot .044+.4747\cdot .0769+.2967\cdot .1231+.044\cdot .1846=.0893$

We consider the four possible cases of the initial draw.

Case 0: No used balls are drawn.

Case 1: 1 used ball is drawn.

Case 2: 2 used balls are drawn.

Case 3: 3 used balls are drawn.

Step 2

We evaluate the probabilities and outcomes of each case.

Case 0:

Case 1:

Case 2:

Case 3:

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.

Step 4

Timothy Wolff

Beginner2022-01-02Added 26 answers

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

$\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7}{{(15\cdot 14\cdot 13)}^{2}}\approx 0.893$ .

Conditionally on this, probability that these 3 balls were not played with because they would have been chosen in the first phase:

Thus, the desired probability is

Vasquez

Expert2022-01-07Added 669 answers

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

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

Now total probability

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