20 numbers chosen at random

20 numbers are chosen at random and independently from numbers 1 to 10 What is the probability that 7 is not chosen?

I thought about 2 ways of solving this.

Number 1: $$P(7 not chosen) =1-P(7 chosen)$$ Now $$P(7 chosen) = \frac{1*\frac{10^{19} }{19}}{\frac{10^{20} }{20}}=\frac{2}{19}$$ So $$P(7 not chosen) =\frac{17}{19}$$ This because we have 20 "places" one is taken by 7 and the other 19 can be filled with any number from 1-10

Number 2: $$P(7 not chosen) =\frac{\frac{9^{20} }{20}}{\frac{10^{20} }{20}}=\frac{9^{20} }{10^{20}}$$ This because for the 20 "places" we can choose any number from 1-10 except 7

But I do not if any of them is correct

The probability that $$7$$ is chosen at any stage is $$1/10$$. Therefore, the probability that $$7$$ is not chosen at any stage is $$9/10$$. You want this to happen for each of the $$20$$ numbers (note that the choices are independent). So this will be $$P(7 \text{ not chosen})^{20}$$. Therefore, it is $$\left(\dfrac{9}{10}\right)^{20} \approx 0.121577= 12.16\%$$