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Let $D$ be the disk with center $(0,0)$ and radius $r$. If $A$ and $B$ are two randomly chosen points in $D$, then what is the probability of the disk $D'$ with center $A$ and radius $AB$ to be contained in $D$? Any hints about this problem?

brick
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1 Answers1

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Assuming that $\|A\|=\rho$, $B$ has to be chosen in a disk centered in $A$ with radius $r-\rho$.

Assuming that $A,B$ are took with respect to a uniform distribution over the unit circle, the pdf of the distance from the origin is supported on $[0,1]$ and given by $f_\rho(x)=2x$, so the wanted probability is given by: $$ \mathbb{P}=\int_{0}^{1}2x(1-x)^2\,dx = \color{red}{\frac{1}{6}}.$$

Jack D'Aurizio
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  • Can you explain exactly which distance is represented by $f_p$? – brick Jun 05 '15 at 20:58
  • @brick: as written, it is not a distance, it is the probability density function of the distance from the origin, for a uniform distribution over the unit disk. – Jack D'Aurizio Jun 05 '15 at 22:27
  • OK, but as far as I know the probability density function of the uniform distribution U[a,b] is $\frac{1}{b-a}$ in the inteval $[a, b]$ and $0$ outside of it. Or I'm wrong? – brick Jun 05 '15 at 22:55
  • Also I don't get where that $(1-x)^2$ came from? I guess that this is the conditional probability of $D' \subseteq D$ if we know that $A$ is at distance $x$ of the center of $D$, but why is it multiplyed by $2x$? Is it like in the discrete case when one can calculate all of the conditional probabilities and then sum them up? – brick Jun 05 '15 at 23:09
  • @brick: what is your knowledge about continuous distributions? Maybe it is not the case to discuss them in all the details here, you can study them by yourself. In any case, yes, $(1-x)^2$ comes from a conditional probability: if $A$ lies at distance $\rho$ from the origin, any $B$-point in a circle with area $\pi(1-\rho)^2$ is "good to go". – Jack D'Aurizio Jun 05 '15 at 23:25
  • As you can see my knowledge is very little. However I think I get the main idea: for every possible distance $x$ you calculate the conditional probability, multiply it by $2x$ (the actual distance) and integrate from $0$ to $1$. Is this a formula which let's you calculate the probability of an event if you know it's conditional probability for every single value of some random variable? – brick Jun 05 '15 at 23:39
  • @brick: informally speaking, your probability is given by the weighted sum of the probabilities that $A$ lies at distance $\rho$ from the origin (that is encoded by the $2x$ in the integral) and $B$ lies at distance at most $1-\rho$ from $A$ (that is encoded by $(1-x)^2$), and we are applying the integral equivalent of $$\mathbb{P}(A)=\sum_{B}\mathbb{P}(B)\cdot \mathbb{P}(A\mid B).$$ – Jack D'Aurizio Jun 06 '15 at 08:33
  • OK, thanks for the explanation. One last question. Shouldn't the encoding of 'A lies at distance $p$ from the origin' be just $1$. I'm asking because, if that distance has uniform distribution in the interval $[0, 1]$, then it's probability density function is $1$ in the interval $[0, 1]$ and $0$ outside of it? – brick Jun 06 '15 at 12:57
  • @Brick: if we take a uniform distribution over a unit ball, the distance from the center is not uniformly distributed, but in any case its density function is not supported outside $[0,1]$. – Jack D'Aurizio Jun 06 '15 at 13:01
  • Oh, I finally understood it completely! If the distance of $A$ to the origin is the random variable $X$, then $F_X (a) = P(X \leq a) = a^2$ for $a \in [0, 1]$, so the probability density function of $X$ is $F_X ' (a) = 2a$. That's why you are using $2x$ in the integral, right? – brick Jun 06 '15 at 14:16