Counting-Probability - permutations
Math Is FUN!!
WE HEAR!!:
Teen mothers who live with their parents are less likely to use marijuana than teen moms in other living arrangements.
WE HEAR!!:
She'll probably take the offer
WE HEAR!!:
The chance of rain tomorrow is 75%
HOW IT IS RELATED TO MATH ??
Math and Probability
- To decide "how likely" an event is, we need to count the number of times an event could occur and compare it to the total number of possible events.
- Such a comparison is called the probability of the particular event occurring.
The Mathmetical theory of counting is known as combinatorial analysis.
FActorial NotatiOn
FActorial nOtaTion USe
- A simple way of writing the product of all the positive whole numbers up to a given number.
What is N?!!
- n factorial is the product of all the integers from 1 to n.
- n factorial is represented with an exclamation mark : n!
- n!=n(n-1)(n-2)...(3)(2)(1)
- EXAMPLE:
- NOTE:
counting: Principles and rules
counting techniques
- Handle large masses of statistical data
- Understanding probability.
counting principle
Number of Outcomes of an Event
- event E defined as E = "season of the year"
- We write the "number of outcomes of event E" as n(E).
So in the example,
n(E)=4,
since there are 4 seasons in a year.30 days
counting rule
Addition Rule
- Let E1 and E2 be mutually exclusive events:
The number of times event E will occur can be given by the expression:
n(E) = n(E1) + n(E2)
where
n(E) = Number of outcomes of event E
n(E1) = Number of outcomes of event E1
n(E2) = Number of outcomes of event E2
example
- Consider a set of numbers
- Let the events E1, E2 and E3 be defined as:
E = choosing a negative or an odd number from S;
E1= choosing a negative number from S;
E2 = choosing an odd number from S.
- Find n(E).
- E1 and E2 are mutually exclusive events
- n(E) = n(E1) + n(E2)
= 2 + 5
= 7
multiplication rule
- Two events E1 and E2 are to be performed and the events E1 and E2 are independent events. (one does not affect the other's outcome)
- Generalization:
Together there will be n(E1) × n(E2) possible outcomes of the two events.
That is, if event E is the event that both E1 and E2 must occur, then
- n(E) = n(E1) × n(E2)
example 1
- Say the only clean clothes you've got are
- Answer:
- We have :
2 t-shirts and with each t-shirt we could pick 4 pairs of jeans. Altogether there are
2×4=8 possible combinations.
- We could write
E1 = "choose t-shirt" and
E2 = "choose jeans"
n(E1) = 2 (since we had 2 t-shirts)
n(E2) = 4 (since there were 4 pairs of jeans)
- So total number of possible outcomes is given by:
n(E) = n(E1) × n(E2) = 2 × 4 = 8
example 2
- What is the total number of possible outcomes when a pair of coins is tossed?
- The events are described as:
E1 = toss first coin (2 outcomes, so n(E1) = 2.)
E2 = toss second coin (2 outcomes, so n(E2) = 2.)
- They are independent, since neither toss affects the outcome of the other toss.
- So n(E) = n(E1) × n(E2) = 2 × 2 = 4
permutations
Theorem 1 - Arranging n Objects
Theorem 2 - Number of Permutations
The number of permutations of n distinct objects taken r at a time, denoted by Pnr, where repetitions are not allowed, is given by
Pnr=n(n−1)(n−2)...(n−r+1)=n!(n−r)!
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!