Binomial distribution

Binomial Distribution

Take an experiment made of several Bernoulli trials - say, tossing a coin repeatedly, with $S$ = success (heads) and $F$ = failure (tails) in each trial. Suppose the coin is tossed 6 times and we want exactly one success among the 6 trials. Listing every arrangement with exactly one $S$ and five $F$'s gives exactly 6 distinct outcomes: $SFFFFF, FSFFFF, FFSFFF, FFFSFF, FFFFSF, FFFFFS$ - one for each position the single $S$ could occupy. Similarly, the number of arrangements with exactly two successes and four failures among 6 trials is $\dfrac{6!}{4!\times2!}=15$. As the number of trials $n$ grows, listing out every arrangement like this becomes impractical, so a general formula is needed instead.

Deriving the distribution for $n=3$ trials. Consider three Bernoulli trials with probability $p$ of success and $q=1-p$ of failure in each. The sample space is $S=\{SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF\}$, and the number of successes, $X$, can take the values $0,1,2,3$. Because the trials are independent, the probability of any particular outcome is the product of the individual trial probabilities:

• $P(X=0)=P(FFF)=q\cdot q\cdot q=q^3$.
• $P(X=1)=P(SFF)+P(FSF)+P(FFS)=pq\cdot q+q\cdot pq+q\cdot qp=3pq^2$ (there are 3 arrangements with exactly one $S$, each with probability $pq^2$).
• $P(X=2)=P(SSF)+P(SFS)+P(FSS)=p\cdot pq+p\cdot qp+q\cdot pp=3p^2q$ (3 arrangements with exactly two $S$'s, each with probability $p^2q$).
• $P(X=3)=P(SSS)=p^3$.

These four probabilities, $q^3,\,3pq^2,\,3p^2q,\,p^3$, are exactly the four terms of the binomial expansion $(q+p)^3=q^3+3q^2p+3qp^2+p^3$, in order: the probability of $x$ successes is the $(x+1)$-th term of $(q+p)^3$. Since $q+p=1$, the four probabilities automatically add up to 1, as any valid probability distribution must.

The general case. For $n$ Bernoulli trials, the same reasoning generalises: the probability of $0,1,2,\ldots,n$ successes turns out to be the $1$st, $2$nd, $3$rd, ..., $(n+1)$-th term of the expansion of $(q+p)^n$. To see why, note that $x$ successes together with $(n-x)$ failures, in some order, is exactly what "$x$ successes in $n$ trials" means. The number of ways to choose which $x$ of the $n$ trial-slots are successes is $\dfrac{n!}{x!\,(n-x)!}={}^{n}C_{x}$. In every one of those arrangements, independence again means the probability is the product of $x$ factors of $p$ (one per success) and $(n-x)$ factors of $q$ (one per failure), i.e. $p^x q^{n-x}$, regardless of which particular arrangement it is. Multiplying the count of arrangements by the probability of each arrangement gives

which is precisely the $(x+1)$-th term of $(q+p)^n$.

Written out for every value of $x$, the probability distribution of the number of successes $X$ in $n$ Bernoulli trials is:

This distribution is called the binomial distribution with parameters $n$ and $p$, because once $n$ and $p$ are fixed, the entire distribution is determined. It is written $X\sim B(n,p)$, read "$X$ follows a binomial distribution with parameters $n,p$". The probability of exactly $x$ successes, $P(X=x)$, is also written $P(x)$ and is given by the probability function (probability mass function, p.m.f.) of the binomial distribution:

A binomial distribution with $n$ Bernoulli trials and success probability $p$ in each trial is denoted $B(n,p)$ or $X\sim B(n,p)$.

Two conditions to keep in mind: (i) the number of trials should be fixed; (ii) the trials should be independent.

Worked examples.

Ex.1 - A fair coin is tossed 10 times, so $X\sim B(10,\tfrac12)$ with $n=10$, $p=q=\tfrac12$, and $P(X=x)={}^{10}C_x(\tfrac12)^x(\tfrac12)^{10-x}$.

(i) Exactly six heads: $P(X=6)={}^{10}C_6(\tfrac12)^{10}=\dfrac{210}{1024}=\dfrac{105}{512}$. …