Explanation for Binomial Distribution in Statistics

Question

In the Binomial Distribution equation for Discrete Probability Statistics, what exactly does the first part, $n!/(x!(n-x)!)$ mean for the equation, and do I have that written out correctly? And overall, what does binomial distribution do for statistics?

Answer 1

Throw a die $6$ times. The number of $1$s you see is one of seven numbers: $0,1,2,3,4,5,6$.

What is the probability that that number is $4$, i.e. you get four $1$s and two non-$1$s?

There are actually $15$ ways that could happen: Using a $0$ to represent any outcome besides a $1$, and $1$ to represent a $1$, we have: \begin{align} & 0 0 1 1 1 1 \\ & 0 1 0 1 1 1 \\ & 0 1 1 0 1 1 \\ & 0 1 1 1 0 1 \\ & 0 1 1 1 1 0 \\ & 1 0 0 1 1 1 \\ & 1 0 1 0 1 1 \\ & 1 0 1 1 0 1 \\ & 1 0 1 1 1 0 \\ & 1 1 0 0 1 1 \\ & 1 1 0 1 0 1 \\ & 1 1 0 1 1 0 \\ & 1 1 1 0 0 1 \\ & 1 1 1 0 1 0 \\ & 1 1 1 1 0 0 \end{align}

Q: How did I know that $15$ is how many there are?

A: $\qquad \dfrac{6!}{4!(6-4)!} = 15$.

Each of these $15$ outcomes has probability $p^4 q^2$, where $p$ is the probability of a $1$ on each trial and $q=1-p$ is the probability of a non-$1$. So the sum of these $15$ probabilities is $$ 15 p^4 q^2 = \frac{6!}{4!(6-4)!} p^4 q^6. $$ That is the probability of getting exactly $4$ successes and $2$ failures in $6$ trials.

Answer 2

You did write it out correctly. That coefficient (called the binomial coefficient) just says how many ways there are to choose x objects out of n objects. So for statistics, if we want to assign equal probability to each of the 2^n events corresponding to n flips of a coin (for example), if we want the probability of x heads, we have to count all of the different ways that can occur, which is the binomial coefficient, and multiply that by the probability of a given series of n coin flips (1/2^n).