I am learning about the binomial coefficient and counting. We define: $$ (1+x)^m = \sum^m_{n=0} \begin{pmatrix} m \\ n \end{pmatrix} x^n $$ the coefficients of the powers of $x^n$ represent the number of ways of choosing $n$ objects from a set of $m$. Is there a way to convert this into a probability distribution by dividing each coefficient by $$ \sum^m_{n=0} \begin{pmatrix} m \\ n \end{pmatrix}. $$


We have a set of three objects, $\{a, b , c\}$, we can draw 1 object three ways $(a + b + c)$, two objects three ways $(ab + ac + bc)$ and three objects one way $(abc)$. Then,

$$ (1+x)^3 = 1x^0 + 3x^1 + 3x^2 + 1x^3 $$ If we divide each coefficient by the sum of coefficients (in this case 8) then they represent probabilities(?). Can we then say that $(1+x)^n$ is the generator for a distribution function $$ p_n = \frac{1}{\sum^m_{n=0} \begin{pmatrix}m \\ n \end{pmatrix}} \begin{pmatrix} m \\ n \end{pmatrix} $$ Such that

$$ (1+x)^n = \sum_n p_n x^n $$


2 Answers 2


Note that $$ p_n =\binom{m}{n}/\sum_{n=0}^{m}\binom{m}{n}=\binom{m}{n}/2^{m}. $$ So $$ \sum_{n=0}^{m} p_n x^n=\frac{(1+x)^m}{2^{m}}=\left(\frac{1}{2}+\frac{x}{2}\right)^m=\sum_{n=0}^m\binom{m}{n}\left(\frac{1}{2}\right)^n \left(\frac{1}{2}\right)^{m-n}x^n.\tag{1} $$ You can think of $p_n$ as the probability of getting $n$ heads when we toss a fair coin $m$ times.

More precisely if $X\sim\text{Binom}(m,1/2)$, i.e. $X$ is a random variable that follows a binomial distribution with $m$ trials and probability of success $1/2$, then $p_n=P(X=n)$.

We say that the series in (1) is the probability generating function of $X$ as it encodes the probabilities of $X$.


What you are observing is on the right track - indeed it makes polynomials exceptionally helpful for combinatorial problems that may be otherwise extremely cumbersome.

Suppose that we are to perform $n$ independent experiments. For each $k$, we have that event $A$ happens with probability $p_k$, and it does not occur with probability $1-p_k := q_k$. Then, consider the product $$(p_1 x + q_1 ) ( p_2 x + q_2 ) \dots (p_n x + q_n )$$ The coefficient of $x^m$ in the above is precisely the probability that $A$ occurs exactly $m$ times in the $n$ experiments.

In your question, you would be computing the case where the $p_k = q_k = 1/2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.