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:
We let
The binomial formula is
The symbol ! means factorial, and . For example,
Also, and by definition.
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
Thus,
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.
NUMBER OF HEADS (r) | |
---|---|
0 | |
1 | |
2 | |
3 | |
4 | |
5 |
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, and 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 section and find the column. In the row where we see 0.0244. Thus, Similarly, and By adding these three probabilities, we have the probability that the number of defects is 3 or more:
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
The expected value and variance for the MSA Electronics example are computed as follows:
Programs 2.2A and 2.2B illustrate how Excel is used for binomial probabilities.