Download Derivation of Binomial Probability Formula

Derivation of Binomial Probability Formula (Probability for Bernoulli Experiments)
One of the most challenging aspects of mathematics is extending knowledge into unfamiliar territory or
unrehearsed exercises. In this category might fall the general concept of “binomial probability,” which
is the blanket under which many mathematical exercises fall. However, using the proper reasoning
techniques, we can devise an elegant and appropriate solution to such exercises.
One classic illustration of binomial probability with which most students are familiar is the case of
determining one’s chances of passing a multiple choice or true/false test on which all answers are mere
guesses. A classic scientific example is that of Gregor Mendel’s famous genetic experiments in which
the colors of peas were examined after various hybridization processes were implemented. For purposes
of developing a formula for binomial probability in this packet, we will consider the following:
“On average, one out of every eight peas is yellow. If two dozen peas are checked at random,
what is the probability that exactly two peas are yellow?”
In this case we are trying to determine a probability associated with the makeup of an entire group being
observed rather than being given explicit knowledge of the makeup of the group and then examining a
subset of it. For this reason, we must employ different techniques from those used in other cases.
For simplicity’s sake, rather than work with a total of 24 peas, let us consider the simpler case of
examining only 5 peas (we will still consider the probability of having exactly 2 yellow peas). The first
issue that comes to mind is that although we know how many peas are yellow, we do not know which
peas are yellow (i.e., the first two, the last two, the second and fourth, etc.). To begin our analysis, let us
work with the case in which the first two peas are yellow and the others are green. (Note: we will
assume that all non-yellow peas are green even though other colors may exist; the important point to
make is that we are concerned with only two actual possibilities of pea colors for our experiment, yellow
or anything-other-than-yellow.) Symbolically, we can represent this situation as follows:
Yellow – Yellow – Green – Green – Green
(abbreviated from this point forward as YYGGG)
To determine the probability of this particular arrangement occurring, we translate the situation into
words as follows:
“What is the probability of five peas chosen at random being chosen in the following order:
yellow and then yellow and then green and then green and then green?”
Grammatically we may have some issues, but mathematically this question bears significant weight. If
we recall that the word “AND” means we multiply probabilities together, we obtain the following
probability for the above situation:
Prof. Fowler
More concisely, we can use exponent notation to write the preceding probability as follows:
( ) ( )
However, that probability is only for the case in which the two yellow peas occur first. Listed below are
all possible locations in which the two yellow peas could occur:
YYGGG, YGYGG, YGGYG, YGGGY, GYYGG, GYGYG, GYGGY, GGYYG, GGYGY, GGGYY
Computing each of these probabilities as was previously done for the first case above and incorporating
the commutative property of multiplication each time, we observe that the probability in each case is the
same. For example, the seventh case, GYGGY, produces a probability as follows:
We might also note that to determine how many ways we can arrange the 2 yellow and 3 green peas
above, we can simply consider how many ways we can arrange the 2 yellow peas out of the total group
of 5 peas since any pea that is not yellow is green and is, thus, implicitly arranged by being placed into
one of the spots not chosen for a yellow pea. Mathematically, this statement translates into the number
of ways to arrange 2 of a total of 5 items if order does not matter, or, symbolically, ( ).
Since the probability requested in our simplified version of the problem is the probability of having two
peas out of 5 be yellow, and those 2 yellow peas could occur in any one of the ten arrangements listed
above, we translate the question posed as follows:
“What is the probability of having the two yellow peas occur in the first or the second or the
third or the fourth or the fifth or the sixth or the seventh or the eighth or the ninth or the tenth
arrangement listed above?”
Again, we may have some grammatical issues with our translation, but the mathematical concept that
applies is the fact that the word “OR” implies addition of probabilities. Since each arrangement has the
same probability, we are employing repeated addition, which is nothing more than multiplication, to
determine a final answer. Thus, we compute the solution as follows:
(
)
In words, we multiplied the number of arrangements of peas that produce our desired makeup of the
group (2 yellow and 3 green) by the probability of one of these arrangements occurring (since they all
have the same probability); symbolically, we have:
( ) ( )
Prof. Fowler
In general, if we are generating a group comprised of some combination of two types of items (i.e., peas
that are either yellow or not yellow), and we use the following variable definitions:
=
=
=
=
total number of items in the group,
number of items in the group of the desired type,
probability that any given item in the group is of the desired type,
probability that any given item in the group is not of the desired type (Note:
),
then the probability of the group having the desired makeup is:
This formula is commonly referred to as the Binomial Probability Formula. If we apply this formula to
the original problem statement on the first page of this packet, we must have the following:
(the total number of peas in the group)
(the number of yellow peas desired)
(the probability that any given pea is yellow)
(the probability that any given pea is not yellow)
So the desired probability is:
( )
( )
which is the desired result. Of course, we could also have obtained this solution by reasoning out the
problem as was done for our simplified example with only 5 peas, but once we have developed the
formula, there is no need to duplicate the effort; we simply substitute the appropriate quantities into the
formula and calculate the result.
Similar techniques can be applied to all experiments for which the following conditions are met:
1.
2.
3.
4.
The experiment has a fixed number of trials.
Each trial must be independent of all other trials.
Each outcome of each trial must be able to be classified as one of only two categories.
The two categories of outcomes and their associated probabilities are the same for each trial.
As long as all the preceding conditions are satisfied, binomial probability techniques can be applied to
efficiently arrive at a solution to the mathematical challenge being considered. Although there may be a
somewhat significant amount of creativity required, it may be possible to categorize a wide variety of
exercises as binomial probability exercises even if at first they do not appear to be so. █
Prof. Fowler