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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?

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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 $0.9$.

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If $X(X+10)>11$, then $X>1$. So the probability is 0.9. No more calculations needed.

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  • $\begingroup$ How did you get 0.9? $\endgroup$ – user286826 Dec 6 '15 at 22:20
  • $\begingroup$ 90% of the interval is allowed for the inequality. For a uniform distribution, that's all you need to know. $\endgroup$ – Paul Dec 6 '15 at 22:21

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