Why is this incorrect for a uniform distribution question

Question

Running times are uniformly distributed between $8.5$ and $10.5$ seconds. The record is $9.9$ seconds. There are eight runners. Determine the probability that the loser will not break the record.

$X\sim U(8.5, 10.5)$

So $P(X>9.9) = 1-P(X≤9.9) = 1 - F(9.9) =1- \frac{9.9-8.5}{10.5-8.15}=0.3$

Why is that wrong?

Answer 1

Hint: Assume $Y_1 < ... < Y_n$ are order statistics for $U(8.5, 10.5)$. Then, your goal is to find the $P(Y_8 > 9)$.

Answer 2

Determine the probability that the loser will not break the record.

The loser is the one with the maximum run time among the 8 runners.

Thus the answer will be

$$\mathbb{P}[\max(X_i)>9.9]=1-\left[\frac{9.9-8.5}{2}\right]^8\approx 0.94$$

This because is well known that the distribution of max of n i.i.d rv's is

$$F_{X_{(n)}}(t)=\left[F_{X}(t)\right]^n$$

Answer 3

Consider $X_1,\dots X_8\sim \mathcal{U}(8.5,10.5)$ iid. Then the probability, that the loser breaks the record is $$ P(\forall i=1,\dots,8: X_i<9.9)=P(X_1<9.9)^8=(1.4/2)^8\approx 0.06$$

Answer 4

It says the loser does not break the record. So I interpret that others may or may not. Here loser is the last person as per your comment.

So the answer should simply be $~1-[P(X\leq 9.9)]^{8}$.

$[P(X\leq 9.9)]^{8}$ is the probability that all of them break the record. Subtracting from $1$ gives you the probability that at least one of them does not break the record and then certainly the loser does not break the record.