There is a box with $5$ balls, all of different colors. $3$ of these balls are drawn and their colors noted, and the balls are then put back. That procedure is then repeated $3$ times. What are the chances that the same set of balls ARE NOT drawn twice or more?
So, $4$ drawings occur where each time $3$ out of $5$ balls are "picked". From what I know, the probability of something not happening is $1 - $(the chance of it happening). So I have been trying to figure out what the chances that the same set of balls are drawn twice or more, but I'm a bit stuck trying to calculate the chance of that.
Best regards
There are $\binom{5}{3} = 10$ ways to select three of the five balls. For each of the ten ways of selecting the four balls, there are nine ways to draw a different set of balls on the second draw, eight ways to draw a set of balls different from the first two draws on the third draw, and seven ways to draw a set of balls different from the first three on the fourth draw. Thus, the probability of obtaining a different set of balls on each of four draws is $$1 \cdot \frac{9}{10} \cdot \frac{8}{10} \cdot \frac{7}{10}$$