Let a random variable X have a uniform distribution on the interval $[0, 10]$.
Find $P(X(X + 10) > 11)$
Since X has a uniform distribution, the pdf of X is
$$
f(x)=\left\{\begin{array}{ccl}c&\text{if}&a\lt x\lt b\\ 0&\text{elsewhere}\end{array}\right.
$$
Where $$c=\frac{1}{b-a} = \frac{1}{10}$$
And $P(X(X + 10) > 11) = P(X^2 + 10X – 11 > 0)$
How do I go further?
Best Answer
I've solved this without pen and paper so I'm not writing it formally but here's the solution: $X^2 + 10X - 11 > 0$ holds true if $X>1$ or $X<-11$.
In the given range that means $10 > X > 1$ which means for 90% of the total possible values of $X$.
As this is a uniform distribution all values of $X$ are equally likely, so there is a 90% probability if you do a random experiment the value of $X$ would be in $(1, 10)$.
So the probability is $0.9$.