TITLE: Find the probability that the largest value $X_{\max}$ is greater than $0.92$
QUESTION [0 upvotes]: Let $X_1, X_2, \ldots , X_n \sim F$ with corresponding pdf $f(x) = 3x^2, \ 0 \leq x \leq 1$. If $ n = 10$ find the probability that the largest value $X_{\max}$ is greater than $0.92.$
To find this I have I was first asked to find the pdf for $X_{\max}$ which I found to
$$
f_{\max}(x) = nF^{n-1}(x)f(x) = n(x^3)^{n-1}3x^2 = 3nx^{3n-1}
$$
and then afterwards calculated
$$
P(X_{\max} \leq 0.92) = 1 - P(X_{\max} > 0.92) = 1 - \int_0^1 30x^{29} \, {\rm d}x = 1-1 = 0
$$
But I guess this is not right. Where did I do wrong?
TIA for any help.

REPLY [1 votes]: You got everything right except the very last part.  How did you conclude $$\Pr[X_{max} > 0.92] = \int_{\color{red}{x=0}}^{1} 30x^{29} \, dx?$$  It needs to be
$$\Pr[X_{max} > 0.92] = \int_{x=0.92}^1 30x^{29} \, dx.$$

REPLY [1 votes]: $P(X_{max} >0.92)=\int_{0.92}^{1} 30x^{2}dx$. You took the integral from $0$ to $1$.
{"set_name": "stack_exchange", "score": 0, "question_id": 4057641}