Probability of picking a different object with replacement A box contains 6 tennis balls. Peter picks two balls at random from the box, plays with them, and returns them to the box. Paul then picks two more balls at random from the box (they can be the same or different from Peter's balls), plays with them, and returns them to the box. Finally, Mary picks two more balls at random and plays with them. What is the probability that every ball picked was played with exactly once?
I'm not really sure how to approach this question because there's nothing distinguishing each tennis ball from each other. Obviously the first step would be picking 2 balls from 6 but what exactly is the sample space here and how would I go approach this problem?
Thanks!
 A: For the all-distinct event to happen,


*

*Peter has ${\large{\binom{6}{2}}}$ valid choices out of ${\large{\binom{6}{2}}}$ total choices.$\\[8pt]$

*Paul has ${\large{\binom{4}{2}}}$ valid choices out of ${\large{\binom{6}{2}}}$ total choices. $\\[8pt]$

*Mary has ${\large{\binom{2}{2}}}$ valid choices out of ${\large{\binom{6}{2}}}$ total choices.


hence, the desired probability is
$$
{\large{\frac{\binom{6}{2}}{\binom{6}{2}}}}
\cdot
{\large{\frac{\binom{4}{2}}{\binom{6}{2}}}}
\cdot
{\large{\frac{\binom{2}{2}}{\binom{6}{2}}}}
=
\frac{15}{15}
\cdot
\frac{6}{15}
\cdot
\frac{1}{15}
=
\frac{2}{75}
$$
As regards the sample space, if you number the balls $1,...6$ with invisible ink (so the players can't see the numbers but you can), then each player draws a $2$-element subset of the set 
$$B=\{1,...,6\}$$
hence we can regard the sample space, $S$ say, as the set of ordered triples of such $2$-element subsets.

As an example, the triple
$$\bigl(\{2,5\},\{1,3\},\{3,4\}\bigr)$$
is a sample point, corresponding to the outcome where


*

*Peter chooses the $2$-element subset $\{2,5\}$ (i.e., balls $2$ and $5$).$\\[4pt]$

*Paul chooses the $2$-element subset $\{1,3\}$ (i.e., balls $1$ and $3$).$\\[4pt]$

*Mary chooses the $2$-element subset $\{3,4\}$ (i.e., balls $3$ and $4$).



Since ${\large{\binom{6}{2}}}=15$, it follows that $|S|=15^3$.
A: To start off all balls in the box have not been played with at all. So any balls that get picked by Peter have now been played with exactly once. When Paul goes to pick balls there are 2 balls that have been played with and 4 that have not. To satisfy the condition of each ball being played with exactly once Paul must pick 2 balls not played with yet. The probability of that occurring would be 4/6 * 3/5. After that successfully occurs there would now be 4 balls played with and 2 not played with, so to meet the condition Mary must pick the 2 balls not played with. The probability of that occurring being 2/6 * 1/5. Therefore the total probability of all this occurring would be 4/6 * 3/5 * 2/6 * 1/5 = 2/75
