Find the value of theta. Let $X$ have the density function $f(x) = \dfrac{3x^2}{\theta^3}$ for $0 < x < \theta$, and $f(x) = 0$ otherwise. If $P\{X > 1\} = 7/8$, find the value of $\theta$.
I don't know

We know that $\Pr(X\le 1)=\dfrac{1}{8}$. But
$$\Pr(X\le 1)=\int_0^1 \frac{3x^2}{\theta^3}\,dx,$$
which is $\dfrac{1}{\theta^3}$.  Thus we need to solve the equation $\dfrac{1}{\theta^3}=\dfrac{1}{8}$.
Remark: Alternately, we could have used the equation
$$\int_1^\theta \frac{2x^2}{\theta^3}\,=\frac{7}{8}.$$
The integral turns out to be $1-\dfrac{1}{\theta^3}$, so we solve $1-\dfrac{1}{\theta^3}=\dfrac{7}{8}$.
