The Binomial Probability Distributions
We now deal with experiments in which there are only two possible outcomes.
Tossing a fair coin is such an example because there are only two
outcomes; heads and tails. Another example is child birth, here there are only
two possible outcomes; male and female. Experiments such as this involve a
binomial random variable. Binomial
random variables have the following characteristics.
Characteristics of a Binomial Random Variable or Experiment
1. There are only two possible outcomes, one called success and the other called failure
2. All trials are identical.
3. The probability of success remains constant on each trial ( no learning)
4. The trials are independent
The Binomial Probability
Formula
In a binomial experiment the probability of getting exactly R successes in N trials is given by the formula:
Example: A coin is tossed three times or three coins are tossed once. Find
the probability of getting exactly two heads.
Solution:
There are only two outcomes for each trial, heads or tails.
Define success as getting a head on any trial.
The number of trials is three.
The probability of success P = .5
The probability of failure (getting a tail) is q=(1 - p) =.5
The number of successes is R=2
The number of combinations of three things taken two at a time is 3
Therefore from the formula:
If all this mathematical exposition frightens you, don't worry, tables have
been developed to take the pain out of the process. Turn to appendix page A15,
the binomial probability tables and find N=3 and R=2 .Reading under P=.5
you find 0.375. Now that is much easier!
The secret to successfully working binomial probability problems is to
identify success. In the previous problem success was getting a head on any
trial. Students often think that success is getting two heads..
Another point to consider is the use of the word exactly. Exactly two heads
means no more than two heads and no less than two heads. Suppose the problem had
been to find the probability of getting two heads ( note exactly was not used).
Now if three heads occur this must be counted because if you got three heads
then you got two heads. So in this case you would sum the cases for R=2 and R=3
resulting in 0.375 + 0.125 = 0.500.
Example:
Suppose you decide to guess on a multiple choice exam. Suppose there are four
questions, each with four suggested responses. Find the probability of getting
get two correct answers.( note, did not use exactly)
Solution:
Define success as guessing correctly on any one question.
The probability of success is 1/4 = 0.25
The number of trials N = 4.
The number of successes R is at least two. (e.g R=2, 3 or 4)
From the table page A15, find N = 4
For R = 2 and P = 0.25, read 0.211
For R = 3 and P = 0.25, read 0.047
For R = 4 and P = 0.25, read 0.004
Summing these three numbers gives 0.262 as the answer.