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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?

Michael Hardy
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T. Smith
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    Please read this related post: http://math.stackexchange.com/questions/838107/relationship-between-binomial-and-bernoulli – heropup Apr 13 '16 at 23:54

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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).

Chill2Macht
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  • Thank you for the help! What is the difference between the three ways to write that first section out? Your explanation was a neat summary - thanks again. :) – T. Smith Apr 15 '16 at 00:46
  • Sorry, can you rephrase the question? The first sentence of my answer? I don't think I understand. – Chill2Macht Apr 15 '16 at 02:51
  • Well, would you mind answering a question I cannot solve in my textbook? I need the solution written out so that I can see where I'm erring. "If we roll a dice 15 times, what is the probability that we will roll a 6 at least 13 times?" Will the probability for this be 1/6? – T. Smith Apr 15 '16 at 21:01
  • Well the only ways that event can happen are: roll a 6 for 13 times, roll a 6 for 14 times, or roll a 6 for 15 times. There are $\binom{15}{13}=\binom{15}{2}=15*(14/2)=105$ ways 13 rolls can happen, $\binom{15}{14}=\binom{15}{1}=15$ ways 14 rolls can happen, and only $\binom{15}{15}=\binom{15}{0}=1$ ways 15 rolls can happen (all rolls are a 6). Each of these outcomes is equally probable, and hence has probability $1/(2^{15})$. Hopefully you see now how using the binomial coefficients to count the various number of ways something can happen is often useful for calculating probabilities. – Chill2Macht Apr 15 '16 at 23:15
  • That was very explanatory. Thank you, I get where I kept missing, I had only thought to calculate the factorial for 13 times. :) – T. Smith Apr 16 '16 at 16:26
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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.

Michael Hardy
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  • Thank you! I appreciate your explanation, it was very useful. I did wonder, though, is the equation at the end the full binomial distribution equation? Are the probability of successes/failures usually written out in exponents, or are they fractions? – T. Smith Apr 15 '16 at 00:49
  • I don't understand the question in your comment. If the random variable $X$ is the number of successes in $n$ independent trials, with probability $p$ of success on each trial, and probability $q=1-p$ of failure on each trial, then we have $$ \Pr(X=x) = \binom n x p^x q^{n-x} = \frac{n!}{x!(n-x)!} p^x q^{n-x}. $$ Are you trying to ask about that? $\qquad$ – Michael Hardy Apr 15 '16 at 15:59
  • Yes, that was my question, just to see it written out. I did wonder, though, how you calculate the (n x) (I don't know if I wrote that out correctly), instead of using the factorial method. I'm sorry if that doesn't make any sense; I am trying to understand statistics while my algebra is a bit rusty...:P – T. Smith Apr 15 '16 at 20:59
  • @T.Smith : If you have $\dbinom{20} 4 = \dfrac{20!}{4!16!}$, you can write $$ \frac{20\times19\times18\times17 \times \overbrace{16\times 15\times 14\times \cdots \times 1}}{4!16!} = \frac{20\times19\times18\times17}{4\times 3\times2\times1} $$ and then cancel every factor in the denominator with something in the numerator. $\qquad$ – Michael Hardy Apr 16 '16 at 01:51
  • So whenever I see the expression (20 4), I need to write it out as a factorial equation? Thank you! – T. Smith Apr 16 '16 at 16:28
  • @T.Smith : You can write $$ \binom{20} 4 = \frac{\overbrace{20\times 19\times 18\times 17}^\text{4 factors}}{\underbrace{4\times 3\times 2 \times 1}\text{4 factors}}. $$ The point of my previous comment was to explain _why you can do that. $\qquad$ – Michael Hardy Apr 16 '16 at 16:30
  • The formula seems incorrect, I believe it should be:

    $$ 15 p^4 q^2 = \frac{6!}{4!(6-4)!} p^4 q^2. $$

     – Giuseppe Romagnuolo Apr 26 '20 at 23:48