Q. 3.29 - A First Course in Probability

Question

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.

Step-by-Step Solution

Answer

 The probability that none  of those balls has ever been used is $.0893$

Step 1: Four Cases

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.  $p_1=\frac{\left(\begin{array}{c}9\\ 2\end{array}\right)\times 6}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

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

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

Step 2: Probabilities and Outcomes

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.

Step 3: Second Draw Probabilities

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

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

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