Buffon's Needle. Probability to intersect the line.

Question

Problem: A table is ruled by equidistant parallel lines, 1 inch apart. A needle of length 1 inch is tossed at random on the table. What is the probability that it intersects a line?

Let's describe the distance from the needle-center to the nearest line by $D$ and the smallest angle between the needle and and that line by $A$ Drawing a figure leads to the conclusion that the needle intersects a line iff $D/\sin{A}\le 1/2.$

The "at random part" is a hint that the random variables $D\sim\text{unif}[0,1/2]$ and $A\sim\text{unif}[0,\pi/2]$ are independent.

Now my book states a corollary:

$$P((X,Y)\in B)=\int_{-\infty}^{\infty}P((x,Y)\in B)f_X(x) \ \text{d}x. \tag{1}$$

I wan't to compute $P\left(D/\sin{A}\le 1/2\right)$ using (1). But the problem is that I don't understand the corollary. What is $B$ in my case? And what is $P((x,Y)\in B)$?

I do understand that they fix $X=x$, and I can do the same by $A=a$ and write the following

$$P\left(\frac{D}{\sin{A}}\le \frac{1}{2}\right)=\int_{-\infty}^\infty P\left(\frac{D}{\sin{a}}\le \frac{1}{2}\right)f_A(a)\ \text{d}a.$$

With the knowledge that $f_A(a)=1/(\pi/2)=2/\pi$, $f_D(d)=2$ and

$$P\left(\frac{D}{\sin{a}}\le \frac{1}{2}\right)=P\left(D\le\frac{\sin{a}}{2}\right)=\int_0^{\sin{a}/2}2\ \text{d}d=\sin{a},$$

I finally obtain

$$P\left(\frac{D}{\sin{A}}\le \frac{1}{2}\right)=\frac{2}{\pi}\int_{0}^{\pi/2}\sin{a}\ \text{d}a=\frac{2}{\pi.}$$

Correct answer but I still don't really know what I'm doing here with the corrollary. My questions are in bold above.

Answer 1

Short answer:

• $B$ in this question would be a set. $B = \{(u,v) : \frac{v}{sin u} \leq \frac{1}{2} \} $.
• It is the probability of $(x,Y)$ being in set $B$.

Long answer:

To understand and appreciate everything in this question, we need to actually understand what P(A) means, what a random variable is, and the shorthand notation that we use.

Formally, you have some underlying outcome space, let's call it $\Omega$. (Think of a dice roll: your outcome set could be represented as $\{1, 2, 3, 4, 5, 6\})$. Now, $\Omega$ has subsets of outcomes that you can observe. (For instance, you can check whether the outcome is even or not. So, an event $A$ may be $A = \{2,4,6\}$.). So, when we write P(A), we are effectively measuring the size of a set. P(A) takes in a set (A) and returns a number between 0 and 1 to indicate the size (or the probability) of it.

Now, random variables are essentially functions that give (real) values to your outcomes. So, a natural random variable for dice rolls would be an "identity random variable" I such that when it takes an input $\omega \in \Omega$, it returns its value: $I(\omega) = \omega$. (here, since our inputs were real numbers to begin with, this mapping is OK). Another set of popular random variables is indicator random variables; these return 0 or 1 depending on some property of $\omega$. (For instance, for dice roll setting, you can have a random variable $E(\omega)$ that returns 1 if $\omega \in \{2,4,6\}$, and 0 otherwise. This random variable "indicates" whether $\omega$ was even or not).

Now, we always use a shorthand notation: $P(E = 1)$ itself would not mean anything; $E$ is a function and $1$ is a number, so "$E = 1$" is just a logical expression, that is true for some inputs, and false for some: it is not a set. We use this notation to indicate $P(\{\omega : E(\omega) = 1\})$, which is indeed well defined as above.

Now, let's go to your question:

• $((X,Y) \in B)$ is again some form of logical expression. In here, we are checking whether $(U, V)$ satisfies a certain inequality, and $B$ is thus the set of $U,V$ that satisfy it. $B = \{(u,v) : \frac{v}{sin u} \leq \frac{1}{2} \} $.
• Now, fix x. Now, you are checking the size of set of all $\omega$s that have $(x, Y(\omega))$ that are in the set $B$.

Answer 2

The fun part of this is apparently that there are different interpretations which can be put on "random" although the one suggested seems reasonable to me. Needless to say, these give different answers.