Counting-Probability - permutations
Math Is FUN!!
Teen mothers who live with their parents are less likely to use marijuana than teen moms in other living arrangements.
She'll probably take the offer
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 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!
counting: Principles and rules
- Handle large masses of statistical data
- Understanding probability.
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,
since there are 4 seasons in a year.30 days
- 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)
n(E) = Number of outcomes of event E
n(E1) = Number of outcomes of event E1
n(E2) = Number of outcomes of event E2
- 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
- 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)
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)
- Say the only clean clothes you've got are
- 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
- 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
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
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