geometric probability --- parallelograms

Question

Inside a rhombus E with sides 10 unit and one interior angle less than 90 degree , there are 2 parallel ( with E ) parallelograms A and B , both can move freely and uniformly inside E but must keep parallel with E in moving . A is with base 8 unit and adjacent side 6 unit ; while B is with base 5 unit and adjacent side 9 unit . If a point is chosen randomly in E , find the probability that the point lies inside A and B at the same time .

Answer 1

We may assume that everything happens in the unit square $[0,1]^2$.

Begin with the following one-dimensional problem: We have a movable subinterval $J\subset[0,1]$ leaving free space of length $0<\ell\leq{1\over2}$, whose position is uniformly distributed within the given limits. Denote by $p(x)$ the probability that the point $x\in[0,1]$ is covered by $J$. Then $p(1-x)=p(x)$, by symmetry. For $0\leq x\leq{1\over2}$ it is easy to see that $$p(x)=\cases{{\displaystyle{x\over\mathstrut \ell}}\qquad&$(0\leq x\leq \ell)\>,$\cr 1&$\bigl(\ell\leq x\leq{1\over2}\bigr)\>.$\cr}$$ If we now have two such intervals $J_1$, $J_2$ such that $\ell_1<\ell_2\leq{1\over2}$, distributed independently, then the probability $q(x)$ that $x$ is covered by both intervals computes to $$q(x)=\cases{{\displaystyle{x^2\over\mathstrut\ell_1\ell_2}}\qquad&$(0\leq x\leq \ell_1)\>,$\cr {\displaystyle{\mathstrut x\over\mathstrut\ell_2}}\qquad&$(\ell_1\leq x\leq \ell_2)\>,$\cr 1&$\bigl(\ell_2\leq x\leq{1\over2}\bigr)\>.$\cr}$$ In the case at hand we have $\ell_1={1\over5}$, $\ell_2={1\over2}$ for the $x$-direction. If we now assume that $x$ is uniformly distributed in $[0,1]$ as well the overall probability $P_x$ that the random point $x$ is covered by both intervals becomes $$P_x=2\int_0^{1/2}p(x)\>dx=2\int_0^{1/5}10x^2\>dx+2\int_{1/5}^{1/2}2x\>dx={71\over150}\ .$$ Similarly for the $y$-direction: Here $\ell_1={1\over10}$, $\ell_2={2\over5}$. The overall probability $P_y$ that a random point $y$ is covered by both intervals becomes $$P_y=2\int_0^{1/10}25x^2\>dx+2\int_{1/10}^{2/5}{5x\over2}\>dx+2\int_{2/5}^{1/2}1\>dx={71\over120}\ .$$ The probability $P$ that a uniformly distributed random point $(x,y)\in[0,1]^2$ is covered by both random rectangles at the same time is therefore given by $$P=P_x\cdot P_y={5041\over18\,000}\doteq0.280056\ ,$$ as indicated by the OP in a comment (now deleted).

Answer 2

We can either impose oblique $(x,y)$ coordinates on the plane of the three paralellograms, with the $x$- and $y$-axes parallel to the sides of all the parallelograms, or we can transform the plane linearly (preserving relative areas) so that the rhombus becomes a square. Either way, it is convenient to place the axes so that the bottom and left sides of the rhombus lie along the $x$- and $y$-axes, respectively.

Taking the random variables $X_A,Y_A$ as the coordinates of the lower left corner of parallelogram $A$, $X_B,Y_B$ as the coordinates of the lower left corner of parallelogram $B$, and $X_P,Y_P$ as the coordinates of a random point $P$ within the rhombus, let the variables have pairwise independent distributions \begin{align} X_A &\sim U(0,2) \\ Y_A &\sim U(0,4) \\ X_B &\sim U(0,5) \\ Y_B &\sim U(0,1) \\ X_P &\sim U(0,10) \\ Y_P &\sim U(0,10) \\ \end{align} where $U(a,b)$ is the uniform distribution on the interval $[a,b]$.

Now, $P$ is (strictly) inside $A \cap B$ if and only if $$\min\{X_A,X_B\} < X_P < \max\{X_A+8,X_B+5\} \tag Q$$ and $$\min\{Y_A,Y_B\} < Y_P < \max\{Y_A+6,Y_B+1\}. \tag R$$ (If a point on the boundary of $A \cap B$ counts as "inside both $A$ and $B$" then change $<$ to $\leq$ in the previous statement.)

Since the variables are pairwise independent, so are the events $Q$ and $R$ described by equations $(Q)$ and $(R)$, respectively, and the answer to the question, $P(Q \cap R)$, obeys the equation $$ P(Q \cap R) = P(Q) P(R).$$ So consider equation $Q$. For given values $X_A = x_A$ and $X_B = x_B$, let $u = \min\{X_A,X_B\}$ and $v = \max\{X_A+8,X_B+5\}$; then $$P(Q \mid X_A = x_A, X_B = x_B) = P(u < X_P < v) = \frac{1}{10}(v - u)$$ (since $0 \leq u \leq v \leq 10$). So the unconditional probability $P(Q)$ is just $frac{1}{10}$ of the mean size of the interval $[\min\{X_A,X_B\},\max\{X_A+8,X_B+5\}]$. I figure the mean of $\max\{X_A+8,X_B+5\} - \min\{X_A,X_B\}$ is $\frac{71}{15}$, so $P(Q) = \frac{71}{150}$, and by similar reasoning $P(R) = \frac{71}{120}$, so I find that $$ P(Q \cap R) = \left(\frac{71}{150}\right) \left(\frac{71}{120}\right) = 5041/18000 \approx 0.2800555\ldots.$$