Probability of Random Selection Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random, one at a time with replacement. What is the probability that the largest number appearing on a selected coupon be 9?   

My attempt: I tried to break down the sum into 7 parts and find the probability in each case. I do not know how to find the probability of choosing n particular choices( all less than or equal to 9) out of m choices remembering the fact that 9 should appear in atleast one of the 7 draws
For a with replacement selection each draw is independent. Consider the $k^{th }$ draw. Any ticket numbered less than $9$ can be chosen, the probability of chosing such a ticket will be $\frac9{15}$, as any ticket has the same probability, $\frac1{15}$, of being chosen. Hence the probability that in all the $7$ draws the tickets chosen are $\le9$, will be $$\left(\frac9{15}\right)^7$$
Similarly, the probability that in all the $7$ draws the tickets chosen are $\le8$, will be $$\left(\frac8{15}\right)^7$$
But see that in the former case the maximum of the numbers drawn will he $\le9$, while in the latter case the maximum of the numbers drawn will be $\le8$. So the probability that the maximum of the numbers drawn is exactly $9$ will be  $$\left(\frac9{15}\right)^7- \left(\frac8{15}\right)^7$$
For a with replacement selection each draw is independent. Consider the $k^{th }$ draw. Any ticket numbered less than $9$ can be chosen, the probability of chosing such a ticket will be $\frac9{15}$, as any ticket has the same probability, $\frac1{15}$, of being chosen. Hence the probability that in all the $7$ draws the tickets chosen are $\le9$, will be $$\left(\frac9{15}\right)^7$$
Similarly, the probability that in all the $7$ draws the tickets chosen are $\le8$, will be $$\left(\frac8{15}\right)^7$$
But see that in the former case the maximum of the numbers drawn will he $\le9$, while in the latter case the maximum of the numbers drawn will be $\le8$. So the probability that the maximum of the numbers drawn is exactly $9$ will be  $$\left(\frac9{15}\right)^7- \left(\frac8{15}\right)^7$$