given $n$ balls that writed about any ball a number $i$, namely, we have $n$ balls: $1,2,3,4,...,n$.
I takes two balls (randomly,independently, without return of the ball that I take).
Will define $X$ - the minimum number of balls I took out.
How can I calculate $P(X)$ ?
I tried in the next way: $P(X=x)=P(X\leq x)-P(X\leq {x-1} )=\frac{x}{n}\cdot \frac{n-x}{n-1}+\frac{n-x}{n} \cdot \frac{x}{n-1}$
And I don't undertstand what is my problem.
The answer is : $\frac{1}{n}\cdot \frac{n-x}{n-1}+\frac{n-x}{n} \cdot \frac{1}{n-1}$
If $F$ denotes the number of the first ball that is taken out and $S$ the number of the second ball then:$$P(X=x)=P(F=x\wedge S>x)+P(F>x\wedge S=x)=$$$$P(F=x)P(S>x\mid F=x)+P(F>x)P(S=x\mid F>x)=\frac1n\frac{n-x}{n-1}+\frac{n-x}n\frac1{n-1}$$