# Pobability of getting a particular number as the largest with replacement

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

My approach:-

I understood that it will be $$9^7-8^7$$ (excluding cases with 8 as maximum) But i was also thinking this way

first i fill 1 place out of 7 by 9 and then fill remaining 6 places with any digit from 1 to 9. So total favourable cases = 7C1 * $$9^6$$.

But these are not equal numerically. What is wrong in 2nd way. Pls clarify.

• Since you are allowing replacement you are badly overcounting. Specifically, any choice that contains $n$ $9's$ is counted $n$ times.
– lulu
Jun 4 '20 at 12:39
• sir i did not understand. Kindly elaborate. It is like 1 nine already present. Then 1-9 present in remaining cases. It ensures atleast 1 nine right. How is it overcounting i am not able to vizualise... In textbook the answer is given as $9^7-8^7$ which i could make out from this. But i did not understand where i went wrong Jun 4 '20 at 12:44
• Consider the selection $9,1,9,9, 2, 7,9$ as an example. You count that once when you choose to put a $9$ in the first slot, and then choose the other six at random. Then you count it again when you choose to put a $9$ in the third slot. And again for the fourth slot, and again for the seventh. So so count this particular choice four times.
– lulu
Jun 4 '20 at 12:47
• If that is still unclear, then I suggest writing it out completely for a smaller sample space. Suppose you are rolling two dice and that you ask for the probability that the greatest roll is a $4$. The first method gives $4^2-3^2=7$. The second gives $2\times 4=8$. Once again, the second method is over counting. Specifically, it counts the pair $(4,4)$ twice.
– lulu
Jun 4 '20 at 12:50
• Oh yes! understood now. Thank you sir :) Jun 4 '20 at 12:54

I don't think you need to worry about overcounting here. Using lulu's example with 2 dice, the probability to get two '4' is $$\frac{1}{36}$$, not $$\frac{2}{36}$$, because in case two '4' are different (e.g. two dice are read and blue and the order matters), the total number of outcomes would be $$2 \times 36 = 72$$ (because for each choice, e.g. 1 and 6, you get 2 sequences: 1R6B, 1B6R).
Another example: you have a two-sided dice with numbers 1 and 0, and you toss them $$n$$ times. If the outcomes with the same number of 0s and 1s is the same, you have a total of $$n+1$$ outcomes (all 0s, all 1s, n-1 0s + one 1, etc). But the probability of getting, for example, all 0s would not be $$\frac{1}{n+1}$$, it would still be $$\frac{1}{2^n}$$.
Applying this logic to your case, the probability to get the sequence with the largest number=9 (imagine you have a 15-sided dice that you toss 7 times) is $$1 - \frac{6^7}{15^7}$$