Find the probability of the event that number $1$ and number $7$ were chosen first and third, respectively at the experiment of choosing five numbers from this set of numbers $\{1,2,3,4,5,6,7,8,9\}$ if:
a) the choices are with replacement and order is count
b) the choices are without replacement and order is count
a) Size of sample space: $n(s)=9^5$ Size of the event $A$: $n(A)=9^3$
$$P(A)=\frac{n(A)}{n(s)}=\frac{9^3}{9^5}=\frac{1}{81}$$
On
If sampling is done with replacement, at any time all choices are equally likely, and we have independence. So the probability $1$ is chosen first is $1/9$. Given that $1$ was chosen first, the probability $7$ is chosen third is $1/9$. Thus the answer is $1/81$.
The situation is different without replacement. All orders of choosing are equally likely. The probability $1$ is chosen first is $1/9$. Given that $1$ was chosen first, the probability $7$ is chosen third is $1/8$, for a probability of $1/72$.
Remark: In the without replacement part, all permutations are equally likely. So first is $1$, third is $7$ has the same probability at, say, sixth is $1$ and fourth is $7$. To answer the question, it is maybe useful for the intuition to imagine that we are calculating the probability that the first is $1$ and the second is $7$.
For a) you are correct. Note that (particularly with replacement) you can neglect all the other picks and just say the probability is (pick 1 first)(pick 7 third)$=(\frac 19)^2=\frac 1{81}$
For b) again you can ignore the other picks-think of swapping the second and third picks. So you have $\frac 19$ to get the pick 1 first and $\frac 18$ to get the pick 7 (second/third) because 1 isn't available any more.