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ELEC 303 – Random Signals. Lecture 4 – Counting Farinaz Koushanfar ECE Dept., Rice University Sept 3, 2009. Lecture outline. Reading: Section 1.6 Review Counting Principle Permutations Combinations Partitions . Independence - summary. Two events A and B are independent if

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elec 303 random signals

ELEC 303 – Random Signals

Lecture 4 – Counting


ECE Dept., Rice University

Sept 3, 2009

lecture outline
Lecture outline
  • Reading: Section 1.6
  • Review
  • Counting
    • Principle
    • Permutations
    • Combinations
    • Partitions
independence summary
Independence - summary

Two events A and B are independent if

P(AB)=P(A).P(B) (Symmetric)

If also, P(B)>0, independence is equivalent to


A and B conditionally independent if given C, P(C)>0, P(AB|C)=P(A|C)

Independence does not imply conditional independence and vice versa

review independent trials and the binomial probabilities
Review: independent trials and the Binomial probabilities

Independent trials: sequence of independent but identical stages

Bernoulli: there are 2 possible results at each stage

Figure courtesy of Bertsekas&Tsitsiklis,

Introduction to Probability, 2008

review bernoulli trial
Review: Bernoulli trial

P(k)=P(k heads in an n-toss sequence)

From the previous page, the probability of any given sequence having k heads: pk(1-p)n-k

The total number of such sequences is

Where we define the notation

counting discrete uniform law
Counting: discrete Uniform law

Let all sample points be equally likely

Counting is not always simple – challenging

Parts of counting has to do with combinatorics

We will learn some simple rules to help us count better

counting principal
Counting principal

r steps

At step i, there are ni choices

Total number of choices n1,n2,…,nr

Figure courtesy of Bertsekas&Tsitsiklis,

Introduction to Probability, 2008

two examples
Two examples

The number of telephone numbers

The number of subsets of an n-element set

Permutations and combinations

k permutations

The order of selection matters!

Assume that we have n distinct objects

k-permutations: find the ways that we can pick k out of these objects and arrange them in a certain order

Permutations: k=n case

Example: Probability that 6 rolls of a six-sided die give different numbers

  • There is no certain order of the events
  • Combinations: number of k-element subsets of a given n-element set
  • Two ways of making an ordered sequence of k-distinct items:
    • Choose the k items one-at-a-time
    • Choose k items then order them (k! possibilities)
    • Thus,
different sampling methods
Different sampling methods
  • Draw k balls from an Urn with n balls
    • Sampling with replacement and ordering
    • Sampling without replacement and ordering
    • Sampling with replacement without ordering
    • Sampling without replacement with ordering

Given an n-element set and nonnegative integers n1,n2,…,nr whose sum is n

How many ways do we have to form the first subset? The second?...

The total number of choices

  • Several terms cancel:
  • Count the number of possible ways for Ace distribution
  • Count the number of ways to
  • Distribute the rest of 48 cards

The 52 cards are dealt to 4 players

Find the probability that each one gets an ace?

Count the size of the sample space

summary of counting results
Summary of counting results

Permutations of n objects: n!

k-Permutations of n objects: n!/(n-k)!

Combinations of k out of n objects:

Partitions of n objects into r groups,

with the ith group having ni objects: