3.4.1 The Binomial Random Variable and Distribution

• In most binomial experiments:
  • The total number of $S$‘s is of interest.
  • More important than knowing which specific trials yielded $S$‘s.

Definition

The binomial random variable $X$ associated with a binomial experiment consisting of $n$ trials is defined as $X$ := the number of $\mathrm{S}$‘s among the $n$ trials

Example

Suppose, for example, that $n=3$ . Then there are eight possible outcomes for the experiment: SSS SSF SFS SFF FSS FSF FFS FFF

From the definition of $X$, $X\left(SSF\right)=2$ , $X\left(SFF\right)=1$, and so on. Possible values for $X$ in an $n$-trial experiment are $x=0,1,2,\dots ,n$ . We will often write $X\sim \mathrm{Bin}\left(n,p\right)$ to indicate that $X$ is a binomial rv based on $n$ trials with success probability $p$ .

NOTATION

Because the pmf of a binomial rv $X$ depends on the two parameters $n$ and $p$ , we denote the pmf by $b\left(x;n,p\right)$ .

Consider first the case $n=4$ for which each outcome, its probability, and corresponding $x$ value are displayed in Table 3.1. For example,

$$\begin{array}{rl}P(SSFS)& =P(S)\cdot P(S)\cdot P(F)\cdot P(S)\text{(independent trials)}\\ & =p\cdot p\cdot (1-p)\cdot p(\text{constant }P(S))\\ & =p^3\cdot (1-p).\end{array}$$

Table 3.1 Outcomes and Probabilities for a Binomial Experiment with Four Trials

Outcome	x	Probability	Outcome	x	Probability
SSSS	4	$p^4$	FSSS	3	$p^3(1-p)$
SSSF	3	$p^3(1-p)$	FSSF	2	$p^2(1-p{)}^2$
SSFS	3	$p^3(1-p)$	FSFS	2	$p^2(1-p{)}^2$
SSFF	2	$p^2(1-p{)}^2$	FSFF	1	$p(1-p{)}^3$
SFSS	3	$p^3(1-p)$	FFSS	2	$p^2(1-p{)}^2$
SFSF	2	$p^2(1-p{)}^2$	FFSF	1	$p(1-p{)}^3$
SFFS	2	$p^2(1-p{)}^2$	FFFS	1	$p(1-p{)}^3$
SFFF	1	$p(1-p{)}^3$	FFFF	0	$(1-p{)}^4$

In this special case, we wish $b\left(x;4,p\right)$ for $x=0,1,2,3$, and 4. For $b\left(3;4,p\right)$ , let’s identify which of the 16 outcomes yield an $x$ value of 3 and sum the probabilities associated with each such outcome:

$$\begin{array}{rl}& b\left(3;4,p\right)\\ =& P\left(FSSS\right)+P\left(SFSS\right)+P\left(SSFS\right)+P\left(SSSF\right)\\ =& 4p^3\left(1-p\right)\end{array}$$

There are four outcomes with $X=3$ and each has probability $p^3\left(1-p\right)$

• the order of $S$ ‘s and $F$ ‘s is not important, only the number of $S$ ‘s,

so $b\left(3;4,p\right)=\left\{\begin{array}{l}\text{ number of outcomes }\\ \text{ with }X=3\end{array}\right\}\cdot \left\{\begin{array}{l}\text{ probability of any particular }\\ \text{ outcome with }X=3\end{array}\right\}$

Similarly, $b\left(2;4,p\right)=6p^2{\left(1-p\right)}^2,$which is also the product of the number of outcomes with $X=2$ and the probability of any such outcome.

In general, $b\left(x;n,p\right)=\left\{\begin{array}{l}\text{ number of sequences of }\\ \text{ length }n\text{ consisting of }x{S}^{\prime }\mathrm{s}\end{array}\right\}\cdot \left\{\begin{array}{l}\text{ probability of any }\\ \text{ particular such sequence }\end{array}\right\}$

Since the ordering of $S$ ‘s and $F$ ‘s is not important, the second factor in the previous equation is $p^x{\left(1-p\right)}^{n-x}$

• e.g., the first $x$ trials resulting in $S$ and the last $n-x$ resulting in $F$.

The first factor is the number of ways of choosing $x$ of the $n$ trials to be $S$‘s

• that is, the number of combinations of size $x$ that can be constructed from $n$ distinct objects (trials here).

Theorem

$b\left(x;n,p\right)=\{\begin{array}{cc}\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}& x=0,1,2,\dots ,n\\ 0& \text{ otherwise }\end{array}$

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