The life time of a light bulb is uniformly distribuited between 500 and 800 hours You have just bought 10 light bulbs, What is the probability that the highest observed life time is greater than 700 hours?
My work: Since $X\sim U(500,800)$, $f_{xn}(x)$ = $\frac{10}{b-a}(\frac{x-a}{b-a})^{9}$
$P(X>700)$ = \begin{equation} \tag{2} \int_{700}^{800} \frac{10}{300}(\frac{x-500}{300})^{9} dx \end{equation} I got at the end $$\frac{1}{300}[\frac{300^{10}}{300^{9}}-\frac{200^{10}}{300^{9}}]$$
so $P(X>700) = 0.9826$ did I make a mistake?
Your answer is correct, but it is much, much simpler to reason as follows.
Let $A$ represent the event that the highest observed lifetime among $10$ light bulbs is greater than $700$ hours. So we want $\Pr[A]$. Let $\bar A$ represent the complementary event, which is that none of the $10$ light bulbs has a lifetime of more than $700$ hours; i.e., all light bulbs fail on or before $700$ hours. Then
$$\Pr[\bar A] = \left(\frac{700 - 500}{800 - 500}\right)^{10} = \left(\frac{2}{3}\right)^{10},$$ hence $$\Pr[A] = 1 - \Pr[\bar A] = 1 - \left(\frac{2}{3}\right)^{10} = \frac{58025}{59049} \approx 0.982658.$$