35. [20] For each $i \in\{1, \ldots, 10\}, a_{i}$ is chosen independently and uniformly at random from $\left[0, i^{2}\right]$. Let $P$ be the probability that $a_{1}<a_{2}<\cdots<a_{10}$. Estimate $P$.
An estimate of $E$ will earn $\left\lfloor 20 \min \left(\frac{E}{P}, \frac{P}{E}\right)\right\rfloor$ points.
Answer: $\quad P \approx 0.003679$
Solution: The probability that $a_{2}>a_{1}$ is $7 / 8$. The probability that $a_{3}>a_{2}$ is $7 / 9$. The probability that $a_{4}>a_{3}$ is $23 / 32$. The probability that $a_{5}>a_{4}$ is $17 / 25$. The probability that $a_{6}>a_{5}$ is $47 / 72$. The probability that $a_{7}>a_{6}$ is $31 / 49$. The probability that $a_{8}>a_{7}$ is $79 / 128$. The probability that $a_{9}>a_{8}$ is $49 / 81$. The probability that $a_{10}>a_{9}$ is $119 / 200$.
Assuming all of these events are independent, you can multiply the probabilities together to get a probability of around 0.05 . However, the true answer should be less because, conditioned on the realization of $a_{1}<a_{2}<\cdots<a_{k}$, the value of $a_{k}$ is on average large for its interval. This makes $a_{k}<a_{k+1}$ less likely. Although this effect is small, when compounded over 9 inequalities we can estimate that it causes the answer to be about $1 / 10$ of the fully independent case.
$P$ was approximated with $10^{9}$ simulations (the answer is given with a standard deviation of about $\left.2 \times 10^{-6}\right)$.