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Two random numbers $x,y\in\mathbb R$ are independently chosen from an interval $[-2,2]$. Evaluate the probability for $x$ and $y$ such that the following is valid: $1\le|x|+|y|\le2$ or $1\le x^2+y^2\le4$.

We can define two separate events:

$A: 1\le|x|+|y|\le2$

$B: 1\le x^2+y^2\le4$

and evaluate $P(A\cup B)$.

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$P(A)=\frac{m(A)}{m(\Omega)},P(B)=\frac{m(B)}{m(\Omega)},P(A\cap B)=P(A)P(B)$$ where $m(\Omega)=16$ is the area of a square with side $a=4$.

Event $A$:

enter image description here

$m(A)$ is the surface of marked area, which can be found by integration: $$4\left(\int_0^2 (2-x)dx-\int_0^1 (1-x)dx\right)=6\Rightarrow P(A)=3/8.$$

Event $B$:

enter image description here

$$m(B)=2\left(\int_{-2}^2 \sqrt{4-x^2}dx - \int_{-1}^1 \sqrt{1-x^2}dx\right)=3\pi\Rightarrow P(B)=3\pi/16$$ $$P(A\cap B)=9\pi/128\Rightarrow P(A\cup B)=\frac{3(16+5\pi)}{128}$$

Could someone check if this is correct?

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    Why take A and B as independent events? – 2017-01-05
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    @user1131274, Because there is "or" in the problem definition. If there was "and", we wouldn't need event "A" and "B" - we would need only one event. – 2017-01-05
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    "or" just means it asks probability for $A\cup B$. It does not say anything about independence of event $A$ and $B$. When you are equating $P(A\cap B)=P(A)P(B)$, you are assuming A and B are independent which is not true. – 2017-01-05
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    @user300045 how did you draw those very nice charts? – 2017-08-24

2 Answers 2

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As I rather like your geometric approach, note that the probability of the union is obtained by first overlapping the figures, and then calculating the resulting area, and dividing this by the total area.

Here is the figure of the area, in purple, you have to calculate

enter image description here

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    How to get the half of the function $y=\sqrt{1-x^2}$? – 2017-01-05
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    You've got to integrate: $\sqrt{1-x^2}-(1-x)$ – 2017-01-05
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    I got $P(A)=3/8,P(B)=3\pi/16,P(A\cap B)=(8-\pi)/16$. The solution is $P(A\cup B)=(2\pi-1)/8$. Could you check if this is correct? – 2017-01-05
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    Also, if $x$ and $y$ are chosen independently, then why events $A,B$ are not independent. In general, how can we determine if events are dependent or independent? In this case, is it becase there is an intersection between events? – 2017-01-05
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You are taking event A and B as independent which is not true. Instead for calculating $P(A\cap B)$ find the area common to both events and divide it by $m(\Omega)$. For completeness $P(A)=3/8$, $P(B)=\frac{3\pi}{16}$, $P(A\cap B)= \frac{(8-\pi)}{16}$ and $P(A\cup B)=\frac{2\pi-1}{8}$ which you can verify by calculating the appropriate areas.

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    Could you explain what exactly do you mean by that they are independent? – 2017-01-05
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    See when you are equating $P(A\cap B)=P(A)P(B)$, you are assuming A and B are independent which is not true. So for calculating $P(A\cap B)$ you have to look for the case when both of them are true, which means you have to find cases which satisfies both A and B – 2017-01-05
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    Then what would be the equation for $P(A\cup B)$? – 2017-01-05
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    Your equation is correct $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ but to calculate $P(A\cap B)$ you have to find area common to both events and divide it by $m(\Omega)$ – 2017-01-05
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    How do we know that events $A,B$ are not independent? Is it because there are intersection between them? – 2017-01-05
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    See we don't know if they are independent or not. To check if they are independent we have to show $P(A\cap B)=P(A)P(B)$, which doesn't make sense in this case as if we have $P(A\cap B)$ we won't bother with independence. So instead of assuming anything, calculate the $P(A\cap B)$ by finding the probability of both happening. And check this out https://en.wikipedia.org/wiki/Independence_(probability_theory) for understanding independence. – 2017-01-05
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    Could you check these results: $P(A)=3/8,P(B)=3\pi/16,P(A\cap B)=(8-\pi)/16$. The final solution is $P(A\cup B)=(2\pi-1)/8$. Is this correct? – 2017-01-05
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    Yes, this is absolutely right. – 2017-01-05