Probability for continuous uniform distribution from $0$ to $10$

Question

Suppose $X$ has a continuous uniform distribution on the interval $[0, 10]$. Calculate: $Pr(X+\frac{10}{X}>7)$.

Answer 1

Note that:

$x+\frac{10}{x}>7$

$x>0$

$x^2+10>7x$

$x^2-7x+10>0$

$0≤x<2$ or $5<x<∞$.

Well, we know your random variable is $[0,10]$, and is uniform along that interval. So, it must be $\frac{2}{10}+\frac{5}{10}=0.7$

Answer 2

One method is to use the hint of @JeppeStigNielsen. Another is to Use the transformation method for PDFs, and then find the distribution of $Y = X + \frac{10}{X}$ and integrating to find $P(Y > 7),$ being careful to note the support of $Y$. Using the transformation method for CDFs would involve expressing the desired probability as $$1 - P(Y \le 7) = 1 - P\left(X + \frac{10}{X} \le 7\right)$$ and solving the RHS to get the answer in terms of $X$, which amounts to the first method for practical purposes.

You have given no clue as to what you are studying at this point, so I don't know what approach the author/instructor has in mind.

Below is a simulation based on a million $X$'s sampled from $Unif(0,10).$ You can use the simulation result (accurate to two or three places) for comparison with your analytic answer from one of these methods.

x = runif(10^6, 0, 10)
y = x + 10/x;
mean(y > 7)
## 0.700893