Counting and Probability
For the following sections on counting, we need a simple way of writing the product of all the positive whole numbers up to a given number. We use factorial notation for this.
Definition of n!
n factorial is defined as the product of all the integers from 1 to n (the order of multiplying does not matter) .
We write "n factorial" with an exclamation mark as follows: n!
n! = (n)(n − 1)(n − 2)...(3)(2)(1)
Basic Principles of Counting
An efficient way of counting is necessary to handle large masses of statistical data (e.g. the level of inventory at the end of a given month, or the number of production runs on a given machine in a 24 hour period, etc.), and for an understanding of probability.
In this section, we shall develop a few counting techniques. Such techniques will enable us to count the following, without having to list all of the items:
the number of ways,
- the number of samples, or
- the number of outcomes
Let E1 and E2 be mutually exclusive events (i.e. there are no common outcomes).
Let event E describe the situation where either event E1 or event E2 will occur.
The number of times event E will occur can be given by the expression:
n(E) = n(E1) + n(E2)
Now consider the case when two events E1 and E2 are to be performed and the events E1 and E2 areindependent events i.e. one does not affect the other's outcome.
Multiplication Rule in General
Suppose that event E1 can result in any one of n(E1) possible outcomes; and for each outcome of the eventE1, there are n(E2) possible outcomes of event E2.
Together there will be n(E1) × n(E2) possible outcomes of the two events.
n(E) = n(E1) × n(E2)That is, if event E is the event that both E1 and E2 must occur, then
How many different combinations can you choose from 2 t-shirts and 4 pairs of jeans?
There are 2 t-shirts and with each t-shirt we could pick 4 pairs of jeans.
2×4=8 possible combinations.
Theorem 1: Arranging n Objects
n distinct objects can be arranged in n! ways.
Theorem 2: Number of Permutations
The number of permutations of n distinct objects taken r at a time, denoted by nPr, where repetitions are not allowed, is: nPr = n(n−1)(n−2)...(n−r+1) = n! ÷ (n−r)!
Example: In how many ways can a supermarket manager display5 brands of cereals in 3 spaces on a shelf?
- 5P3 = 5! ÷ [(5-3)!] = 5! ÷ 2! = 60
Theorem 3: Permutations of Different Kinds of Objects
The number of different permutations of n objects of which n1 are of one kind, n2 are of a second kind, ... nk are of a k-th kind is: n! ÷ (n1! × n2! × n3! ×... nk!)
Theorem 4: Arranging Objects in a Circle
n distinct objects in a circle can be arranged in (n−1)! ways
Number of Combinations:
The number of combinations in which r objects can be selected from a set of n objects, where repetition is not allowed, is denoted by: nCr = n! ÷ [r! (n - r)!]
Suppose an event E can happen in r ways out of a total of n possible equally likely ways.
Then the probability of occurrence of the event is denoted by: P(E) = r ÷ n
The probability of non-occurrence of the event is denoted by: P(É) = 1 - (r ÷ n)
Therefore, P(E) + P(É) = 1 (the sum of probabilities in any experiment is 1)
Using Sample Spaces:
When an experiment is performed, a sample space of all possible outcomes is set up.
In a sample of N equally likely outcomes, a chance of 1N is assigned to each outcome.
The probability of an event for such a sample is the number of outcomes favorable to E divided by the total number of equally likely outcomes in the sample space S of the experiment.
P(E)=n(E) ÷ n(S)
n(E) is the number of outcomes favorable to E
n(S) is the total number of equally likely outcomes in the sample space S of the experiment.
Properties of Probability:
- 0 ≤ P(event) ≤ 1
- P(impossible event) = 0
- P(certain event) = 1
Example: A card is drawn at random from a deck of cards. What is the probability of getting the 3 of diamond?
Let E be the event "getting the 3 of diamond". An examination of the sample space shows that there is one "3 of diamond" so that n(E) = 1 and n(S) = 52. Hence the probability of event E occuring is given by P(E) = 1 / 52