Question

A needle of lengthlis dropped at random onto a sheet of paper ruled with parallel lines a distancelapart. What is the probability that the needle will cross a line?

Short answer

The probability that the needle will cross a line is $2/\pi$ which is equal to $0.63662$.

Step-by-step solution

Define the Schrodinger equation

• A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field.
• Its answer is related to a particle's probability density in space and time.

Determine the probability of x

The probability of crossing is

$P=P\left(y\right)P(x)$

Where y and x are some parameters that we will use.

Suppose that the distance between the end of the needle that have a hole and the the closest line to it to bey , so that yhave the interval between 0 and 1 (i.e., l ), and the projection along some direction x is in the interval between -l and l (i.e., $-l\le x\le l$).

The condition of crossing a line that is above is $x+y\ge l$(i.e., $x\ge l-y$).

Where the condition of crossing line that is below is $x+y\le 0$(i.e., $x\le -y$).

So, for a given value if y the probability of crossing (by make usage of problem 1.12 is

role="math" localid="1658552457032" $$\begin{gathered} P\left(x\right)={\int }_{-l}^{-y}p\left(x\right)dx+{\int }_{l-y}^{l}p\left(x\right)dx \\ ={\int }_{-l}^{-y}\frac{1}{\pi \sqrt{l^2-x^2}}dx+{\int }_{l-y}^l\frac{1}{\pi \sqrt{l^2-x^2}}dx \\ =\frac{1}{\pi }\left[{\sin }^{-1}{\overline{)\left(\frac{x}{l}\right)}}_{-l}^{-y}+{\sin }^{-1}{\overline{)\left(\frac{x}{l}\right)}}_{l-y}^l\right] \\ =\frac{1}{\pi }\left[-{\sin }^{-1}\left(\frac{y}{l}\right)+2{\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(\frac{l-y}{l}\right)\right] \\ P\left(x\right)=\frac{1}{\pi }\left[\pi -{\sin }^{-1}\left(\frac{y}{l}\right)-{\sin }^{-1}\left(\frac{l-y}{l}\right)\right]...\left(1\right) \end{gathered}$$

Determine the probability of the needle cross the line

Now, using the normalizing condition we can find $\rho \left(y\right)$, where it is equal to $1/l$ because all the values of y are equally likely.

Thus eq. (1) become,

localid="1658553112340" $$\begin{gathered} P_{\mathrm{crossing}}=\frac{1}{\pi }{\int }_0^l\left[\pi -{\sin }^{-1}\left(\frac{y}{l}\right)-{\sin }^{-1}\left(\frac{l-y}{l}\right)\right]dy \\ =\frac{1}{\pi }{\int }_0^l\left[\pi -{\sin }^{-1}\left(\frac{y}{l}\right)\right]dy \end{gathered}$$

To integrate the second term of the integrand we use integration by parts, so we get

localid="1658552944578" $$\begin{gathered} P_{cros\sin g}=\frac{1}{\pi l}{\left[\pi l-2\left\{y{\sin }^{-1}\left(\frac{y}{l}\right)+l\sqrt{1-\frac{y^2}{l^2}}\right\}\right]}_0^l \\ =1-1+\frac{2}{\pi } \\ =\frac{2}{\pi } \end{gathered}$$

Therefore, the probability that the needle will cross a line is $2/\pi$which is equal to (approximately) $0.63662$.