For independent and identical $V_1,V_2\in U(-1,1)$, what is the probability that $V_1^2+V_2^2<1$?
I tried but can't get an answer, the answer is $\frac{\pi}{4}$
$\begingroup$
$\endgroup$
12
-
$\begingroup$ Are the $V_1$ and $V_2$ independent? What have you tried? $\endgroup$– mpiktasCommented May 5, 2015 at 10:25
-
$\begingroup$ @mpiktas yes! they are independent, actually I can't find Idea. $\endgroup$– Hemant RupaniCommented May 5, 2015 at 10:28
-
2$\begingroup$ I suggest that you draw the curve $V_1^2+V_2^2=1$ and the curve $|V_1|+|V_2|=1$ ;-) $\endgroup$– Tommy LCommented May 5, 2015 at 10:32
-
6$\begingroup$ This is trivial - draw a picture, write down the answer. $\endgroup$– Glen_bCommented May 5, 2015 at 10:57
-
4$\begingroup$ $$\iint_{x^2+y^2 \leq 1} \frac 14\,\mathrm dx\,\mathrm dy = \int_{r=0}^1\int_{\theta=0}^{2\pi} \frac 14 r\,\mathrm d\theta\,\mathrm dr$$ Can you take it from here? $\endgroup$– Dilip SarwateCommented May 5, 2015 at 11:25
|
Show 7 more comments
1 Answer
$\begingroup$
$\endgroup$
1
Just draw a graph of it; replace V1 and V2 with x and y respectively. You should have a circle with side length 2 centred about the origin. Then draw a graph of x^2 + y^2 = 1, which is a unit circle centred about the origin. The answer would be the area of that circle divided by the area of the square.
-
3$\begingroup$ That does not seem to make sense to me. How do you draw a circle of side 2? Do you mean a square? $\endgroup$– mdeweyCommented Jan 5, 2018 at 13:35