# 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 *E*1, *E*2 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 *E*1 can result in any one of *n*(*E*1) possible outcomes; and for each outcome of the occurrence *E*1, there are *n*(*E*2) possible outcomes of occurrence *E*2. So if we assign *E* as the occurrence that both *E*1 and *E*2 must occur, then

*n*(*E*) = *n*(*E*1) × *n*(*E*2)

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:

*n**P**r*=*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)! = *4**P**2= *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*!*n*1!×*n*2!×*n*3×...×*n**k*!

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 |

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)