Three people each choose five distinct numbers at random from the numbers 1, 2, . . . , 25, independently of each other.

Question

What is the probability that the three choices of five numbers have no number in common?

I know I have to use inclusion-exlcuion here and that using the compliment is probably the best way to solve this. I can't really think of what the events should be though.

My first thought is to let $A_{i}$ = the event that all three pick the number $i$. This would be a pretty tedious inclusion-exclusion problem though so I was wondering if this is the right set of events to use for this problem or if not, what the right set of events is.

EDIT I know the answer is 0.8109

EDIT I came up with an answer that gets me to the answer in the back of the book and I am 99% sure it's right $$ \sum_{i = 1}^{5} (-1)^{i+1} \cdot \begin{pmatrix} 25 \\ i \end{pmatrix} \cdot \frac{\begin{pmatrix} 25 - i \\ 5 - i \end{pmatrix}^{3}}{\begin{pmatrix} 25 \\ 5 \end{pmatrix}^{3}} $$

Answer 1

Well you don't have to use "inclusion-exclusion" principle to solve this. Just use the fact that, $Probabbility=\frac{Total\ accepted\ outcomes}{Total\ outcomes}$.

The total accepted outcomes turns out to be: $(^{25}_5)(^{20}_5)(^{15}_5)$. And,

Total outcome will be: $\{(^{25}_5)\}^3$.

Thus your answer is, $Probabbility=\frac{(^{25}_5)(^{20}_5)(^{15}_5)}{\{(^{25}_5)\}^3}$

Answer 2

The second person has ${25\choose5}$ ways of picking five numbers. ${20\choose5}$ of these will have no number in common with the first person. The third person also has ${25\choose5}$ ways of picking five numbers. ${15\choose5}$ of these will have no number in common with either of the first two. So the required prob is $$\frac{{20\choose5}{15\choose5}}{{25\choose5}^2}\approx1.6\%$$.

Answer 3

$\frac{{25\choose 5}{20\choose 5} {15 \choose 5}}{{25 \choose 5}^3}$ is the probability that there is no number that is chosen by two or more people

$1-\sum_{i=1}^{5}(-1)^{i+1}{25\choose i} \frac{{25-i\choose 5-i}^3}{{25\choose 5}^3}$ is the probability that there is no number that is chosen by each of the three people