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2.6 The Binomial Distribution

Many business experiments can be characterized by the Bernoulli process. The probability of obtaining specific outcomes in a Bernoulli process is described by the binomial probability distribution. In order to be a Bernoulli process, an experiment must have the following characteristics:

Each trial in a Bernoulli process has only two possible outcomes. These are typically called a success and a failure, although examples might be yes or no, heads or tails, pass or fail, defective or good, and so on.
The probability stays the same from one trial to the next.
The trials are statistically independent.
The number of trials is a positive integer.

A common example of this process is tossing a coin.

The binomial distribution is used to find the probability of a specific number of successes out of n trials of a Bernoulli process. To find this probability, it is necessary to know the following:

\begin{array}{l} n & = & the number of trials \\ p & = & the probability of a success on any single trial \end{array}

$\begin{array}{l} n & = & the number of trials \\ p & = & the probability of a success on any single trial \end{array}$

We let

\begin{array}{l} r & = & the number of successes \\ q & = & 1 - p = the probability of a failure \end{array}

$\begin{array}{l} r & = & the number of successes \\ q & = & 1 - p = the probability of a failure \end{array}$

The binomial formula is

Probability of r successes in n trials = \frac{n!}{r! (n - r)!} p^{r} q^{n - r}

$Probability of r successes in n trials = \frac{n!}{r! (n - r)!} p^{r} q^{n - r}$ (2-9)

The symbol ! means factorial, and $n! = n (n - 1) (n - 2) \dots (1)$ $n! = n (n - 1) (n - 2) \dots (1)$ . For example,

4! = (4) (3) (2) (1) = 24

$4! = (4) (3) (2) (1) = 24$

Also, $1! = 1,$ $1! = 1,$ and $0! = 1$ $0! = 1$ by definition.

Solving Problems with the Binomial Formula

A common example of a binomial distribution is the tossing of a coin and counting the number of heads. For example, if we wished to find the probability of 4 heads in 5 tosses of a coin, we would have

n = 5, r = 4, p = 0.5, and q = 1 - 0.5 = 0.5

$n = 5, r = 4, p = 0.5, and q = 1 - 0.5 = 0.5$

Thus,

\begin{array}{l} P (4 successes in 5 trials) & = & \frac{5!}{4!(5 - 4)!} 0 {.5}^{4} 0 {.5}^{5 - 4} \\ = & \frac{5(4)(3)(2)(1)}{4(3)(2)(1)(1!)} (0 .0625)(0 .5) = 0 .15625 \end{array}

$\begin{array}{l} P (4 successes in 5 trials) & = & \frac{5!}{4!(5 - 4)!} 0 {.5}^{4} 0 {.5}^{5 - 4} \\ = & \frac{5(4)(3)(2)(1)}{4(3)(2)(1)(1!)} (0 .0625)(0 .5) = 0 .15625 \end{array}$

Thus, the probability of 4 heads in 5 tosses of a coin is 0.15625, or about 16%.

Using Equation 2-9, it is also possible to find the entire probability distribution (all the possible values for r and the corresponding probabilities) for a binomial experiment. The probability distribution for the number of heads in 5 tosses of a fair coin is shown in Table 2.8 and then graphed in Figure 2.6.

Table 2.8 Binomial Probability Distribution for n $=$ $=$ 5 and p $=$ $=$ 0.50

NUMBER OF HEADS (r)	$P R O B A B I L I T Y = \frac{5!}{r! (5 - r)!} {(0.5)}^{r} {(0.5)}^{5 − r}$ $P R O B A B I L I T Y = \frac{5!}{r! (5 - r)!} {(0.5)}^{r} {(0.5)}^{5 − r}$
0	$0.03125 = \frac{5!}{0! (5 - 0)!} {(0.5)}^{0} {(0.5)}^{5 - 0}$ $0.03125 = \frac{5!}{0! (5 - 0)!} {(0.5)}^{0} {(0.5)}^{5 - 0}$
1	$0.15625 = \frac{5!}{1! (5 - 1)!} {(0.5)}^{1} {(0.5)}^{5 - 1}$ $0.15625 = \frac{5!}{1! (5 - 1)!} {(0.5)}^{1} {(0.5)}^{5 - 1}$
2	$0.31250 = \frac{5!}{2! (5 - 2)!} {(0.5)}^{2} {(0.5)}^{5 - 2}$ $0.31250 = \frac{5!}{2! (5 - 2)!} {(0.5)}^{2} {(0.5)}^{5 - 2}$
3	$0.31250 = \frac{5!}{3! (5 - 3)!} {(0.5)}^{3} {(0.5)}^{5 - 3}$ $0.31250 = \frac{5!}{3! (5 - 3)!} {(0.5)}^{3} {(0.5)}^{5 - 3}$
4	$0.15625 = \frac{5!}{4! (5 - 4)!} {(0.5)}^{4} {(0.5)}^{5 - 4}$ $0.15625 = \frac{5!}{4! (5 - 4)!} {(0.5)}^{4} {(0.5)}^{5 - 4}$
5	$0.03125 = \frac{5!}{5! (5 - 5)!} {(0.5)}^{5} {(0.5)}^{5 - 5}$ $0.03125 = \frac{5!}{5! (5 - 5)!} {(0.5)}^{5} {(0.5)}^{5 - 5}$

A bar graph illustrating the probability of different numbers of success, ranging from one to six. — Figure 2.6 Binomial Probability Distribution for n $=$ $=$ 5 and p $=$ $=$ 0.50

Figure 2.6 Full Alternative Text

Solving Problems with Binomial Tables

MSA Electronics is experimenting with the manufacture of a new type of transistor that is very difficult to mass-produce at an acceptable quality level. Every hour a supervisor takes a random sample of 5 transistors produced on the assembly line. The probability that any one transistor is defective is considered to be 0.15. MSA wants to know the probability of finding 3, 4, or 5 defectives if the true percentage defective is 15%.

For this problem, $n = 5, p = 0.15,$ $n = 5, p = 0.15,$ and $r = 3, 4,$ $r = 3, 4,$ or 5. Although we could use the formula for each of these values, it is easier to use binomial tables for this. Appendix B gives a binomial table for a broad range of values for n, r, and p. A portion of this appendix is shown in Table 2.9. To find these probabilities, we look through the $n = 5$ $n = 5$ section and find the $p = 0.15$ $p = 0.15$ column. In the row where $r = 3,$ $r = 3,$ we see 0.0244. Thus, $P (r = 3) = 0.0244.$ $P (r = 3) = 0.0244.$ Similarly, $P (r = 4) = 0.0022,$ $P (r = 4) = 0.0022,$ and $P (r = 5) = 0.0001.$ $P (r = 5) = 0.0001.$ By adding these three probabilities, we have the probability that the number of defects is 3 or more:

\begin{array}{l} P (3 or more defects) & = & P (3) + P (4) + P (5) \\ = & 0 .0244 + 0 .0022 + 0 .0001 = 0 .0267 \end{array}

$\begin{array}{l} P (3 or more defects) & = & P (3) + P (4) + P (5) \\ = & 0 .0244 + 0 .0022 + 0 .0001 = 0 .0267 \end{array}$

Table 2.9 A Sample Table for the Binomial Distribution

		P
n	r	0.05	0.10	0.15	0.20	0.25	0.30	0.35	0.40	0.45	0.50
1	0	0.9500	0.9000	0.8500	0.8000	0.7500	0.7000	0.6500	0.6000	0.5500	0.5000
	1	0.0500	0.1000	0.1500	0.2000	0.2500	0.3000	0.3500	0.4000	0.4500	0.5000
2	0	0.9025	0.8100	0.7225	0.6400	0.5625	0.4900	0.4225	0.3600	0.3025	0.2500
	1	0.0950	0.1800	0.2500	0.3200	0.3750	0.4200	0.4550	0.4800	0.4950	0.5000
	2	0.0025	0.0100	0.0225	0.0400	0.0625	0.0900	0.1225	0.1600	0.2025	0.2500
3	0	0.8574	0.7290	0.6141	0.5120	0.4219	0.3430	0.2746	0.2160	0.1664	0.1250
	1	0.1354	0.2430	0.3251	0.3840	0.4219	0.4410	0.4436	0.4320	0.4084	0.3750
	2	0.0071	0.0270	0.0574	0.0960	0.1406	0.1890	0.2389	0.2880	0.3341	0.3750
	3	0.0001	0.0010	0.0034	0.0080	0.0156	0.0270	0.0429	0.0640	0.0911	0.1250
4	0	0.8145	0.6561	0.5220	0.4096	0.3164	0.2401	0.1785	0.1296	0.0915	0.0625
	1	0.1715	0.2916	0.3685	0.4096	0.4219	0.4116	0.3845	0.3456	0.2995	0.2500
	2	0.0135	0.0486	0.0975	0.1536	0.2109	0.2646	0.3105	0.3456	0.3675	0.3750
	3	0.0005	0.0036	0.0115	0.0256	0.0469	0.0756	0.1115	0.1536	0.2005	0.2500
	4	0.0000	0.0001	0.0005	0.0016	0.0039	0.0081	0.0150	0.0256	0.0410	0.0625
5	0	0.7738	0.5905	0.4437	0.3277	0.2373	0.1681	0.1160	0.0778	0.0503	0.0313
	1	0.2036	0.3281	0.3915	0.4096	0.3955	0.3602	0.3124	0.2592	0.2059	0.1563
	2	0.0214	0.0729	0.1382	0.2048	0.2637	0.3087	0.3364	0.3456	0.3369	0.3125
	3	0.0011	0.0081	0.0244	0.0512	0.0879	0.1323	0.1811	0.2304	0.2757	0.3125
	4	0.0000	0.0005	0.0022	0.0064	0.0146	0.0284	0.0488	0.0768	0.1128	0.1563
	5	0.0000	0.0000	0.0001	0.0003	0.0010	0.0024	0.0053	0.0102	0.0185	0.0313
6	0	0.7351	0.5314	0.3771	0.2621	0.1780	0.1176	0.0754	0.0467	0.0277	0.0156
	1	0.2321	0.3543	0.3993	0.3932	0.3560	0.3025	0.2437	0.1866	0.1359	0.0938
	2	0.0305	0.0984	0.1762	0.2458	0.2966	0.3241	0.3280	0.3110	0.2780	0.2344
	3	0.0021	0.0146	0.0415	0.0819	0.1318	0.1852	0.2355	0.2765	0.3032	0.3125
	4	0.0001	0.0012	0.0055	0.0154	0.0330	0.0595	0.0951	0.1382	0.1861	0.2344
	5	0.0000	0.0001	0.0004	0.0015	0.0044	0.0102	0.0205	0.0369	0.0609	0.0938
	6	0.0000	0.0000	0.0000	0.0001	0.0002	0.0007	0.0018	0.0041	0.0083	0.0156

The expected value (or mean) and the variance of a binomial random variable may be easily found. These are

Expected value (mean) = n p

$Expected value (mean) = n p$ (2-10)

Variance = n p (1 - p)

$Variance = n p (1 - p)$ (2-11)

The expected value and variance for the MSA Electronics example are computed as follows:

\begin{array}{l} Expected value & = & n p = 510 .152 = 0 .75 \\ Variance & = & n p (1 - p) = 5(0 .15)(0 .85) = 0 .6375 \end{array}

$\begin{array}{l} Expected value & = & n p = 510 .152 = 0 .75 \\ Variance & = & n p (1 - p) = 5(0 .15)(0 .85) = 0 .6375 \end{array}$

Programs 2.2A and 2.2B illustrate how Excel is used for binomial probabilities.

Screenshot showing the Excel output for binomial distribution calculations. — Program 2.2A Excel Output for the Binomial Example

Figure 2.2A Full Alternative Text

Screenshot showing the formulas behind the cumulative probability and probability of exactly r in the prior output calculations. — Program 2.2B Function in an Excel 2016 Spreadsheet for Binomial Probabilities

Figure 2.2B Full Alternative Text

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Table of Contents for 2.6 The Binomial Distribution

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Table of Contents for
2.6 The Binomial Distribution