skilpadw3

Answered

2022-07-23

Probability with elements of geometry

We have a wire of length 20. We bend this wire in random point. And then bend again to get rectangual frame. What's the probability that area of this rectangle is less than 21.

We have a wire of length 20. We bend this wire in random point. And then bend again to get rectangual frame. What's the probability that area of this rectangle is less than 21.

Answer & Explanation

Reinfarktq6

Expert

2022-07-24Added 18 answers

Step 1

Let $X\sim \mathsf{U}(0,20)$ be the point at which we first bend the pole. Then the height of the rectangle is $H=X\wedge 20-X$ and the width of the rectangle is $W=2(10-H)$, so the area of the rectangle is $A=WH=2H(10-H)$. The distribution of H is U(0,10) by symmetry. To compute $\mathbb{P}(A\u2a7d21)=\mathbb{P}(2H(10-H)\u2a7d21)=\mathbb{P}(H(10-H)\u2a7d\frac{21}{2}),$ consider the function $a(h)=h(10-h)$. We have ${a}^{\prime}(h)=10-2h$, so the maximum of a is 25, attained at $h=5$, with a′ negative on $(0,5)\cup (5,10)$. The equation $h(10-h)=\frac{21}{2}$ has solutions $h=5\pm \frac{\sqrt{58}}{2}$, so

$h(10-h)\u2a7d\frac{21}{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}h\in [0,5-\frac{\sqrt{58}}{2}]\cup [5+\frac{\sqrt{58}}{2},10].$

Step 2

By symmetry we have $\begin{array}{rl}\mathbb{P}(H(10-H)\u2a7d\frac{21}{2}& =2\mathbb{P}(H\u2a7d5-\frac{\sqrt{58}}{2})\\ & =2\cdot \frac{1}{10}(5-\frac{\sqrt{58}}{2})\\ & =1-\frac{\sqrt{58}}{10}.\end{array}$

Let $X\sim \mathsf{U}(0,20)$ be the point at which we first bend the pole. Then the height of the rectangle is $H=X\wedge 20-X$ and the width of the rectangle is $W=2(10-H)$, so the area of the rectangle is $A=WH=2H(10-H)$. The distribution of H is U(0,10) by symmetry. To compute $\mathbb{P}(A\u2a7d21)=\mathbb{P}(2H(10-H)\u2a7d21)=\mathbb{P}(H(10-H)\u2a7d\frac{21}{2}),$ consider the function $a(h)=h(10-h)$. We have ${a}^{\prime}(h)=10-2h$, so the maximum of a is 25, attained at $h=5$, with a′ negative on $(0,5)\cup (5,10)$. The equation $h(10-h)=\frac{21}{2}$ has solutions $h=5\pm \frac{\sqrt{58}}{2}$, so

$h(10-h)\u2a7d\frac{21}{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}h\in [0,5-\frac{\sqrt{58}}{2}]\cup [5+\frac{\sqrt{58}}{2},10].$

Step 2

By symmetry we have $\begin{array}{rl}\mathbb{P}(H(10-H)\u2a7d\frac{21}{2}& =2\mathbb{P}(H\u2a7d5-\frac{\sqrt{58}}{2})\\ & =2\cdot \frac{1}{10}(5-\frac{\sqrt{58}}{2})\\ & =1-\frac{\sqrt{58}}{10}.\end{array}$

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