Probabilty fun

The number of ways to choose $9$ is $1$ . The number of ways to choose $25$ is $1$ . The number of ways to choose the middle number $m$ is $24-9 = 15$ . Therefore, the total number of ways to choose $(9, m, 25)$ is $N_{9,m,25} = 3 \times 2 \times 15$ .

The total number of ways to choose three numbers randomly, $N = 30 \times 29 \times 28$ .

Therefore, the probability of choosing $(9, m, 25)$ , $p = \dfrac {N_{9,m,25}}N = \dfrac {3\times 2\times 15}{30 \times 29 \times 28} = \boxed{\frac 3{812}}$ .