$P\left(X+\frac{10}{X}>7\right)$ of a uniform distribution

Question

Problem: $X$ has a continuous uniform distribution on $[0,10]$. Find $P\left( X + \frac{ 10 } { X } >7\right)$.

So far, I have the PDF $f(x) = 1/10$ and CDF $F(x) = x/10$ for $0 < x < 10$.

I was wondering how to transform this information to help me solve the above probability. How do I solve a probability function containing a random variable?

Answer 1

Hint: $$x+\dfrac{10}{x}-7>0~~\implies~~x^2-7x+10=\left(x-\frac{7}{2}\right)^2-\frac{9}{4}>0$$ Added: You're missing part of the solution. $$\begin{align*}P\left(X+\frac{10}{X}>7\right)&=P\left(\left(X-\frac{7}{2}\right)^2>\frac{9}{4}\right)\\[1ex]&=P(0<X<2)+P(X>5)\\[1ex]&=F(2)-F(0)+1-F(5)\end{align*}$$