A table is ruled with equidistant parallel lines that are unit distance apart. A needle of unit length is randomly thrown on the table. What is the probability that the needle will intersect one of the lines? (4
Answers
Answer:
Let x be the distance from the center of the needle to the closest parallel line, and let θ be the acute angle between the needle and one of the parallel lines.
The uniform probability density function of x between 0 and t /2 is
{\displaystyle {\begin{cases}{\frac {2}{t}}&:\ 0\leq x\leq {\frac {t}{2}}\\0&:{\text{elsewhere.}}\end{cases}}}
\begin{cases}
\frac{2}{t} &:\ 0 \le x \le \frac{t}{2}\\
0 &: \text{elsewhere.}
\end{cases}
Here, x = 0 represents a needle that is centered directly on a line, and x = t/2 represents a needle that is perfectly centered between two lines. The uniform PDF assumes the needle is equally likely to fall anywhere in this range, but could not fall outside of it.
The uniform probability density function of θ between 0 and π/2 is
{\displaystyle {\begin{cases}{\frac {2}{\pi }}&:\ 0\leq \theta \leq {\frac {\pi }{2}}\\0&:{\text{elsewhere.}}\end{cases}}}
\begin{cases}
\frac{2}{\pi} &:\ 0 \le \theta \le \frac{\pi}{2}\\
0 &: \text{elsewhere.}
\end{cases}
Here, θ = 0 radians represents a needle that is parallel to the marked lines, and θ = π/2 radians represents a needle that is perpendicular to the marked lines. Any angle within this range is assumed an equally likely outcome.
The two random variables, x and θ, are independent, so the joint probability density function is the product
{\displaystyle {\begin{cases}{\frac {4}{t\pi }}&:\ 0\leq x\leq {\frac {t}{2}},\ 0\leq \theta \leq {\frac {\pi }{2}}\\0&:{\text{elsewhere.}}\end{cases}}}
\begin{cases}
\frac{4}{t\pi} &:\ 0 \le x \le \frac{t}{2}, \ 0 \le \theta \le \frac{\pi}{2}\\
0 &: \text{elsewhere.}
\end{cases}
The needle crosses a line if
{\displaystyle x\leq {\frac {l}{2}}\sin \theta .}x \le \frac{l}{2}\sin\theta.
Now there are two cases.
Case 1: Short needle Edit
Integrating the joint probability density function gives the probability that the needle will cross a line:
{\displaystyle P=\int _{\theta =0}^{\frac {\pi }{2}}\int _{x=0}^{(l/2)\sin \theta }{\frac {4}{t\pi }}\,dx\,d\theta ={\frac {2l}{t\pi }}.}{\displaystyle P=\int _{\theta =0}^{\frac {\pi }{2}}\int _{x=0}^{(l/2)\sin \theta }{\frac {4}{t\pi }}\,dx\,d\theta ={\frac {2l}{t\pi }}.}
Case 2: Long needle Edit
Suppose {\displaystyle l>t}l > t. In this case, integrating the joint probability density function, we obtain:
{\displaystyle \int _{\theta =0}^{\frac {\pi }{2}}\int _{x=0}^{m(\theta )}{\frac {4}{t\pi }}\,dx\,d\theta ,}\int_{\theta=0}^{\frac{\pi}{2}} \int_{x=0}^{m(\theta)} \frac{4}{t\pi}\,dx\,d\theta ,
where {\displaystyle m(\theta )}m(\theta) is the minimum between {\displaystyle (l/2)\sin \theta }(l/2)\sin\theta and {\displaystyle t/2}t/2 .
Thus, performing the above integration, we see that, when {\displaystyle t<l}t < l, the probability that the needle will cross a line is
{\displaystyle {\frac {2l}{t\pi }}-{\frac {2}{t\pi }}\left\{{\sqrt {l^{2}-t^{2}}}+t\sin ^{-1}\left({\frac {t}{l}}\right)\right\}+1}\frac{2 l}{t\pi} - \frac{2}{t\pi}\left\{\sqrt{l^2 - t^2} + t\sin^{-1}\left(\frac{t}{l}\right)\right\}+1
or
{\displaystyle {\frac {2}{\pi }}\cos ^{-1}{\frac {t}{l}}+{\frac {2}{\pi }}{\frac {l}{t}}\left\{1-{\sqrt {1-\left({\frac {t}{l}}\right)^{2}}}\right\}.} \frac{2}{\pi} \cos^{-1}\frac{t}{l} + \frac{2}{\pi} \frac{l}{t} \left\{1 - \sqrt{1 - \left( \frac{t}{l} \right)^2 } \right\}.
Let x be the distance from the center of the needle to the closest parallel line, and let θ be the acute angle between the needle and one of the parallel lines.
The uniform probability density function of x between 0 and t /2 is
{\displaystyle {\begin{cases}{\frac {2}{t}}&:\ 0\leq x\leq {\frac {t}{2}}\\0&:{\text{elsewhere.}}\end{cases}}}
\begin{cases}
\frac{2}{t} &:\ 0 \le x \le \frac{t}{2}\\
0 &: \text{elsewhere.}
\end{cases}
Here, x = 0 represents a needle that is centered directly on a line, and x = t/2 represents a needle that is perfectly centered between two lines. The uniform PDF assumes the needle is equally likely to fall anywhere in this range, but could not fall outside of it.
The uniform probability density function of θ between 0 and π/2 is
{\displaystyle {\begin{cases}{\frac {2}{\pi }}&:\ 0\leq \theta \leq {\frac {\pi }{2}}\\0&:{\text{elsewhere.}}\end{cases}}}
\begin{cases}
\frac{2}{\pi} &:\ 0 \le \theta \le \frac{\pi}{2}\\
0 &: \text{elsewhere.}
\end{cases}
Here, θ = 0 radians represents a needle that is parallel to the marked lines, and θ = π/2 radians represents a needle that is perpendicular to the marked lines. Any angle within this range is assumed an equally likely outcome.
The two random variables, x and θ, are independent, so the joint probability density function is the product
{\displaystyle {\begin{cases}{\frac {4}{t\pi }}&:\ 0\leq x\leq {\frac {t}{2}},\ 0\leq \theta \leq {\frac {\pi }{2}}\\0&:{\text{elsewhere.}}\end{cases}}}
\begin{cases}
\frac{4}{t\pi} &:\ 0 \le x \le \frac{t}{2}, \ 0 \le \theta \le \frac{\pi}{2}\\
0 &: \text{elsewhere.}
\end{cases}
The needle crosses a line if
{\displaystyle x\leq {\frac {l}{2}}\sin \theta .}x \le \frac{l}{2}\sin\theta.
Now there are two cases.
Case 1: Short needle Edit
Integrating the joint probability density function gives the probability that the needle will cross a line:
{\displaystyle P=\int _{\theta =0}^{\frac {\pi }{2}}\int _{x=0}^{(l/2)\sin \theta }{\frac {4}{t\pi }}\,dx\,d\theta ={\frac {2l}{t\pi }}.}{\displaystyle P=\int _{\theta =0}^{\frac {\pi }{2}}\int _{x=0}^{(l/2)\sin \theta }{\frac {4}{t\pi }}\,dx\,d\theta ={\frac {2l}{t\pi }}.}
Case 2: Long needle Edit
Suppose {\displaystyle l>t}l > t. In this case, integrating the joint probability density function, we obtain:
{\displaystyle \int _{\theta =0}^{\frac {\pi }{2}}\int _{x=0}^{m(\theta )}{\frac {4}{t\pi }}\,dx\,d\theta ,}\int_{\theta=0}^{\frac{\pi}{2}} \int_{x=0}^{m(\theta)} \frac{4}{t\pi}\,dx\,d\theta ,
where {\displaystyle m(\theta )}m(\theta) is the minimum between {\displaystyle (l/2)\sin \theta }(l/2)\sin\theta and {\displaystyle t/2}t/2 .
Thus, performing the above integration, we see that, when {\displaystyle t<l}t < l, the probability that the needle will cross a line is
{\displaystyle {\frac {2l}{t\pi }}-{\frac {2}{t\pi }}\left\{{\sqrt {l^{2}-t^{2}}}+t\sin ^{-1}\left({\frac {t}{l}}\right)\right\}+1}\frac{2 l}{t\pi} - \frac{2}{t\pi}\left\{\sqrt{l^2 - t^2} + t\sin^{-1}\left(\frac{t}{l}\right)\right\}+1
or
{\displaystyle {\frac {2}{\pi }}\cos ^{-1}{\frac {t}{l}}+{\frac {2}{\pi }}{\frac {l}{t}}\left\{1-{\sqrt {1-\left({\frac {t}{l}}\right)^{2}}}\right\}.} \frac{2}{\pi} \cos^{-1}\frac{t}{l} + \frac{2}{\pi} \frac{l}{t} \left\{1 - \sqrt{1 - \left( \frac{t}{l} \right)^2 } \right\}.