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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

Best Answer 

 #1
avatar+25750 
+4

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
 #1
avatar+25750 
+4
Best Answer

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

heureka Aug 5, 2020
 #3
avatar+118665 
0

That's pretty neat, heureka!!!!

 

I like this problem, even though I would have no clue as to the  solution!!!!

 

 

cool cool cool

CPhill  Feb 20, 2021
 #2
avatar+113244 
+1

Thanks Heureka,

I would have done this with a probability contour map.     

 Aug 5, 2020

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