A Peek into Probability
A Pre-lesson look at Probability and Counting
Why would mathmaticians study the concept of probability ?
Multiplication
Consider the case when two occurrences E1, E2 are to be performed and that they are independent occurrences (one outcome doesn't affect the other's outcome)
(An example is the jeans, t-shirts and heels idea shown in the video E1 being jeans, E2 t-shirts, and E3 heels)
General rule:
Suppose that occurrence E1 can result in any one of n(E1) possible outcomes; and for each outcome of the occurrence E1, there are n(E2) possible outcomes of occurrence E2. So if we assign E as the occurrence that both E1 and E2 must occur, then
n(E) = n(E1) × n(E2)
Multiplication shows us the many possibilities there are when dealing with 2 or more different occurrences.
Factorial Notation:
n! = (n)(n − 1)…(1)
NOTE: YOU CAN’T SIMPLIFY A FRACTION OF FACTORIALS. ex: ( 4!/ 2! ≠ 2!)
Ex. 4!/ 2! = 24/ 2 =12
Permutations with identical elements:
Concept A)
In general, n distinct elements can be arranged in n! ways.
Concept B)
The number of permutations of n distinct elements in E taken r elements of E at a time, denoted by nPr, without repetitions, is represented as:
nPr=n(n−1)(n−2)...(n−r+1)=n!/(n−r)!
EX. How many words can be formed from the letters of my first name "LANA"?
4! / (4-2)! = 4P2= 12
Concept C)
The number of different permutations of n elements of which n1 and n2 are two different kinds of elements, with nk being the k-th kind of element, is found as such:
n!n1!×n2!×n3×...×nk!
Concept D)
The number of permutations where repetition is considered of a set E consisting of n elements is equal to n!
For example, let's say we invite five dinner guests but unfortunately our dinner table only seats three.
we choose to seat three guests (r=3) among five total guests (n=5). We find the number of permutations (shown in the figure) so the in this case, there are sixty permutations. (5P3 =5! / (5-3)! = 60)
simple huh? :D and easily applicable to day to day life