Probability that $U_1 \geq U$, $U_2 \geq U$, $U_3 < U$, $U_4 \geq U$, $U_5 < U$, $U_6 \geq U$, $U_7 \geq U$, for i.i.d. uniform $U_k$s and $U$ Let $U,U_1,U_2,...$ be independant, on [0,1] uniform distributed random variables.
Let $E$ := {$U_1 \geq U,U_2 \geq U,U_3 < U,U_4 \geq U, U_5 < U,U_6 \geq U,U_7 \geq U$}.
Find the probabiliy $P(E)$ and then find the probability $P(U_8 \geq U |E)$.

I am struggling with this task and therefore need help, thanks!
 A: Recall that
$$\mathbb{P}[U \le u] = \max\{\min\{u,1\},0\})$$
Observe that


*

*$$\mathbb{P}[\max\{U_3, U_5\} \le u] = (\max\{\min\{u,1\},0\})^2$$

*$$u = \max\{\min\{u,1\},0\}, u \in [0,1]$$

*$$\mathbb{P}[\min\{U_1, U_2, U_4, U_6, U_7\} \le u] = 1 - (1 - \mathbb{P}[U_1 \le u])^5$$

*$$E = \{\max\{U_3, U_5\} < U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$
$$ = \{\max\{U_3, U_5\} \color{red}{\le} U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$
$$ = \{\max\{U_3, U_5\} \le U \} \cap \{U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}$$


*$$P(\{\max\{U_3, U_5\} \le U \}) = P(\{\max\{U_3, U_5\} - U \le 0\})$$


$$ = \int_0^1 \int_0^u f_{\max\{U_3, U_5\},U} du_{3,5}du$$
$$ = \int_0^1 \int_0^u f_{\max\{U_3, U_5\}}f_{U} du_{3,5}du$$
$$ = \int_0^1 \int_0^u f_{\max\{U_3, U_5\}}(1) du_{3,5}du$$
$$ = \int_0^1 \int_0^u (2u_{3,5})(1) du_{3,5}du$$
$$= 1/3$$
Or simply note that
$$P(\{\max\{U_3, U_5\} - U \le 0\}) = P(\{\max\{U_3, U_5, U\} = U\})$$
where the RHS is 1/3 by symmetry


*$$P(\{U \le \min\{U_1, U_2, U_4, U_6, U_7\} \}) = P(\{U = \min\{U, U_1, U_2, U_4, U_6, U_7\} \}) = 1/6$$


Integrate that if you want. :P


*So how does 4 relate to 5 and 6 in terms of computing $P(E)$?

