Terms
 Sample Space: The set of all possible values of a random variable is called as sample space
 Examples: Sample space for
 die roll S = {1,2,3,4,5,6}
 Coin Flip S = { H, T }
 Examples: Sample space for
 Cardinality of Set: The size of the elements in a set is called as cardinality
Probability

How likely an event is to occur

Probability of an event is a number that is
 A value between 0 and 1
 O corresponds to "unlikely" or almost never
 1 correspons to "almost surely"

Represent
P(event) = value

Probability is sometime expressed as percentage (0% and 100%)

Rules of Probabilty
 Probability of event A is greater than or equal to zero
P(A) >= 0
 Probability of Sample Space is one
P(S) = 1
 If A1,A2,A3 … are the sequence of mutually exclusive events
P(A1 U A2 U A3 U....) = U A(i) = Σ P(A ) i
 Probability of event A is greater than or equal to zero

Examples: Consider a Coin Flip
 S = { H, T}
 P(H) = P(T) = 1/2
 P(H and tail) = 0
 P(H or T) = P(S) = 1

Discrete Vs Continuous Probability
 Discrete = finite set of unique values
 Coin toss, die roll, number of cars in house hold
 P(E) = 1/N where N is number of possible Events
 Continuous = infinite set of values
 Person height, amount of rainfall in a day
 Instead, better to measure intervals
 Discrete = finite set of unique values

Counting Methods
 Counting Simple Points
 Consider a coin Flip S = {H, T}
 Flip a coin 3 Three times (Sampling with replacement) How many different arrangements are possible?
 Direct approach: list all the arrangements and count them
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
 How many arrangements begin or end with a HEAD
 S(H??) U (S ??H) = HHH, HHT, HTH, HTT, THH, TTH = 6
 Multiplication Rule
 We need to travel from A to C through B
 Three Ways from A to B
AB = {AB(1), AB(2), AB(3) }
 Two ways to get from B to C
BC = { BC(1), BC(2)}
 How many ways of getting from A to C
 Clearly number of options is #AB * #BC => 3*2 = 6
 Counting Simple Points

Example: A container has three balls: Red, Blue, Yellow. You draw each of the balls in turn without replacement. How many possible ways/arrangements are there
 First draw => 3 possible options
 Second draw => 2 possible options (one is already draws)
 Third draw => 1 possible option
 Total number of possible draws is => 3 * 2 * 1 => 6

Permuation:
 Given a set of elements all distinct arrangement s of the elementes are called the permuatation of a set
 The number of ways you can arrange N elements of N possibilies
P(n,n) => N * N1 * N2 ....* 2* 1 = N!
 Suppose we have 5 balls (RGBWO) and you draw 3 balls without replacement. How many possible arrangments are there
5 * 4 * 3 => 60 5 * 4 * 3 * 2 * 1 5!  =  2*1 (53)!
 Number of possible arrangements with K elements out of n elements
n! P(n,k)=  nk! Where P represent Permuations
 Example:
 Number of unique 5 letter words that can be arranged from letters {a,b,c,d,e} => 5^5 (5Pow(5)) => 3125
 Number of unique letters if each letter appears only once => 5! => 120
 Number of 3letter words, each letter repeats only once => P(5,3) => 5!/(53)! => 60

Combinatorics:
 Combinatorics is a branch of mathematics concerning the study of finite or countable discrete structures
 Given a bag of balls B = {R,G,B,W,O}
 If we pull out 2 balls the number of arrangements => 5!/3! = 20
 What if we don’t care about order i.e => {R,B} and {B,R} to be same pick, How can we calcualte the number of unique picks
 The number of picks including duplicates => P(5,2).
 Each pick has 2! possible arrangments
 C (n,k) = P(n,k)/k! = n!/(k! * (nk)!)
 C(5,2) => 5!/ (3!2!) => (120 / 6 2) => 10

Binomial Coeffecients
 Expressed as
 Example: You toss a coin 10 times and record the result, What is probability of getting exactly 4 heads
 Total number of => POW(2,10)
 Each arrangement is a choice to get 4 heads amount 10 tosses
C(10,4)
 P(4 heads) = (C(10,4))/POW(2,10) => 0.205
 Example: You toss a coin 10 times and record the result, What is probability of getting exactly 4 heads or less
 The value is 0.377

Example: A class has 15 girls and 30 boys. Pick 10 children at random. What the probability that you will pick exactly 3 girls
 answer:

Multinomial Coeffecients:
 Examples: 10 students need to be formed into 3 groups consisting of 4,3 and 3 members respectively, In how many ways can be student be assigned to these groups
 First => C(10,4)
 Second => C(6,3)
 Third => C(3,3) => 1
 Number of ways = C(10,4)* C(6,3)
 In general, number of arrangements of n elements into k groups of size n = {n1,n2,n3 …..}
 By applying the above formuala => 10!/(4! * 3! * 3!)
 Examples: 10 students need to be formed into 3 groups consisting of 4,3 and 3 members respectively, In how many ways can be student be assigned to these groups