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A stick has a length of 5 units. The stick is then broken at two points, chosen at random. What is the probability that all three resulting pieces are longer than 1 unit?

 Aug 5, 2020
 #1
avatar+25543 
+2

A stick has a length of \(5\) units. The stick is then broken at two points, chosen at random.

What is the probability that all three resulting pieces are longer than \(1\) unit?

 

 

The probability that all three resulting pieces are longer than \(1\) unit = \(\dfrac{ A_{\color{darkorange}orange} } {A} \)

\(\begin{array}{|rcll|} \hline H^2 + \left(\dfrac{S}{2}\right)^2 &=& S^2 \\ H^2 &=& S^2 -\dfrac{S^2}{4} \\ H^2 &=& \dfrac{3S^2}{4} \qquad \mathbf{H=\dfrac{S\sqrt{3}}{2}} \qquad (1) \\ S^2 &=& \dfrac{4H^2}{3} \\ S &=& \dfrac{2H}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\ \mathbf{S} &=& \mathbf{\dfrac{2H\sqrt{3}}{3}} \qquad (2) \\ \hline S &=& \dfrac{2H\sqrt{3}}{3} \quad | \quad H = 5 \\ \mathbf{S} &=& \mathbf{\dfrac{10\sqrt{3}}{3}} \\ \hline 2A &=& H*S \quad | \quad H=5,\ \mathbf{S=\dfrac{10\sqrt{3}}{3}} \\ \mathbf{2A} &=& \mathbf{\dfrac{50\sqrt{3}}{3}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \tan(30^\circ) &=& \dfrac{1}{x} \quad | \quad \tan(30^\circ) = \dfrac{1}{\sqrt{3}}\\ \dfrac{1}{\sqrt{3}} &=& \dfrac{1}{x} \\ \mathbf{ x } &=& \mathbf{\sqrt{3}} \\ \hline \mathbf{s} &=& \mathbf{S-2x} \quad | \quad \mathbf{S=\dfrac{10\sqrt{3}}{3}},\ \mathbf{ x =\sqrt{3}} \\ s &=& \dfrac{10\sqrt{3}}{3}-2\sqrt{3} \\ s &=& \dfrac{10\sqrt{3}}{3}-\dfrac{6\sqrt{3}}{3} \\ \mathbf{ s } &=& \mathbf{\dfrac{4\sqrt{3}}{3}} \\ \hline \mathbf{h^2 + \left(\dfrac{s}{2}\right)^2} &=& \mathbf{s^2} \\ h^2 &=& s^2 -\dfrac{s^2}{4} \\ h^2 &=& \dfrac{3s^2}{4} \\ \mathbf{h} &=& \mathbf{\dfrac{s\sqrt{3}}{2}}\quad | \quad \mathbf{ s =\dfrac{4\sqrt{3}}{3}} \\ h &=& \mathbf{\dfrac{\dfrac{4\sqrt{3}}{3}\sqrt{3}}{2}} \\ h &=& \dfrac{4}{2} \\ \mathbf{h} &=& \mathbf{2} \\ \hline 2A_{\color{darkorange}orange}&=& h*s \quad | \quad h=2,\ \mathbf{ s=\dfrac{4\sqrt{3}}{3}} \\ 2A_{\color{darkorange}orange} &=& 2*\dfrac{4\sqrt{3}}{3} \\ \mathbf{2A_{\color{darkorange}orange}} &=& \mathbf{\dfrac{8\sqrt{3}}{3}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{ A_{\color{darkorange}orange} } {A} &=& \dfrac{ 2A_{\color{darkorange}orange} } {2A} \\\\ &=& \dfrac{ \dfrac{8\sqrt{3}}{3} } {\dfrac{50\sqrt{3}}{3}} \\\\ &=& \dfrac{8} {50} \\\\ \mathbf{ \dfrac{ A_{\color{darkorange}orange} } {A} } &=& \mathbf{ \dfrac{4}{25} } \\ \hline \end{array}\)

 

 

laugh

 Aug 5, 2020
 #2
avatar+110713 
+1

Thanks Heureka,

I would have done this with a probability contour map.     

 Aug 5, 2020

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