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The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Review of Exam I
Sections 2.2 -- 4.5
Jiaping Wang
Department of Mathematical Science
02/18/2013, Monday
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Outline Sample Space and Events
Definition of Probability
Counting Rules
Conditional Probability and Independence
Probability Distribution and Expected Values
Bernoulli, Binomial and Geometric Distributions
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Definition 2.1A sample space S is a set that includes all possible
outcomes for a random experimentlisted in a mutually exclusive and exhaustive way.Mutually Exclusive means the outcomes of the set do
not overlap.Exhaustive means the list contains all possible
outcomes. Definition 2.2:
An event is any subset of a sample space.
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There are three operators between events:Intersection: ∩ --- A∩B or AB – a new event consisting of common
elements from A and BUnion: U --- AUB – a new event consisting of all outcomes from A or B.Complement: ¯, A, -- a subset of all outcomes in S that are not in A.
Event Operators and Venn Diagram
AUB A∩B A
S SS
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Commutative laws:
Associate laws:
Distributive laws:
DeMorgan’s laws:
Some Laws
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Suppose that a random experiment has associated with a sample space S. A probability is a numerically valued function that assigned a number P(A) to every event A so that the following axioms hold:
(1) P(A) ≥ 0
(2) P(S) = 1
(3) If A1, A2, … is a sequence of mutually exclusive events (that is Ai∩Aj=ø for any i≠j), then
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2. 0≤ P(A) ≤1for any event A.
3. P(AUB) = P(A) + P(B) if A and B are mutually exclusively.
Some Basic Properties
1. P( ø ) = 0, P(S) = 1.
5. If A is a subset of B, then P(A) ≤ P(B).
4. P(AUB) = P(A) + P(B) – P(A∩B) for general events A and B.
6. P(A) = 1 – P(A).
7. P(A∩B) = P(A) – P(A∩B).
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Theorem 2.1. For events A1, A2, …, An from the sample space S,
We can use induction to prove this.
Inclusive-Exclusive Principle
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Determine the Probability Values
The definition of probability only tells us the axioms that the probability function must obey; it doesn’t tell us what values to assign to specific event.
For example, if a die is balanced, then we may think P(Ai)=1/6 for Ai={ i }, i = 1, 2, 3, 4, 5, 6
The value of the probability is usually based on empirical evidence or on careful thought about the experiment.
However, if a die is not balanced, to determine the probability, we need run lots of experiments to find the frequencies for each outcome.
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Fundamental Principle of Counting:If the first task of an experiment can result in n1 possible
outcomes and for each such outcome, the second task can result in n2 possible outcomes, then there are n1n2 possible outcomes for the two tasks together.
Theorem 2.2
The principle can extend to more tasks in a sequence.
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Order Is Important Order Is Not Important
With Replacement nr Crn+r-1
Without Replacement Prn Cr
n
Order and Replacement
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Theorem 2.5 Partitions
Consider a case: If we roll a die for 12 times, how many possible ways to have2 1’s, 2 2’s, 3 3’s, 2 4’s, 2 5’s and 1 6’s? Solution: First, choose 2 1’s from 12 which gives 12!/(2!10!), second, since there aretwo positions are filled by 1’s, the next choice appears in the left 10 positions, so there are 10!/(8!2!) ways, and so similar for next other selections which provides the final result is 12!/(2!10!)x10!/(2!8!)x8!/(3!5!)x5!/(2!3!)x3!/(2!1!)x1!/(1!0!)
=12!/(2!x2!x3!x2!x2!x1!)
Theorem 2.5 Partitions. The number of partitioning n distinct objects into k groups containing n1, n2,•••, nk objects, respectively, is
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Definition 3.1
If A and B are any two events, then the conditional probability of A given B, denoted as P(A|B), is
Provided that P(B)>0.
Notice that P(A∩B) = P(A|B)P(B) or P(A∩B) = P(B|A)P(A).
This definition also follows the three axioms of probability.
(1) A∩B is a subset of B, so P(A∩B )≤P(B), then 0≤P(A|B)≤1;(2) P(S|B)=P(S∩B)/P(B)=P(B)/P(B)=1;(3) If A1, A2, …, are mutually exclusively, then so are A1∩B, A2 ∩B, …; and
P(UAi|B) = P((UAi) ∩B)/P(B)=P(U(Ai ∩B)/P(B)=∑P(Ai ∩B)/P(B)= ∑P(Ai|B).
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Theorem 3.2: Multiplicative Rule. If A and B are any two events, then
P(A∩B) = P(A)P(B|A)= P(B)P(A|B)
If A and B are independent, thenP(A∩B) = P(A)P(B).
Definition 3.2 and Theorem 3.2
Definition 3.2: Two events A and B are said to be independent if
P(A∩B)=P(A)P(B).This is equivalent to stating that
P(A|B)=P(A), P(B|A)=P(B)If the conditional probability exist.
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Theorem of Total Probability:If B1, B2, …, Bk is a collection of mutually exclusive and
exhaustive events, then for any event A, we have
Bayes’ Rule. If the events B1, B2, …, Bk form a partition of the sample space S, and A is any event in S, then
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A random variable X is said to be discrete if it can take on only a finite number – or a countably infinite number – of possible values x. The probability function of X, denoted by p(x), assigns probability to each value x of X so that the following conditions hold:
1. P(X=x)=p(x)≥0;2. ∑ P(X=x) =1, where the sum is over all
possible values of x.
A random variable is a real-valued function whose domain is a sample space.
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The distribution function F(b) for a random variable X isF(b)=P(X ≤ b);
If X is discrete,
Where p(x) is the probability function. The distribution function is often called the cumulative
distribution function (CDF).
Any function satisfies the following 4 properties is a distribution function:
1. 2.
3. The distribution function is a non-decreasing function: if a<b, then F(a)≤ F(b). The distribution function can remain constant, but it can’t decrease as we increase from a to b.
4. The distribution function is right-hand continuous:
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Definition 4.4 The expected value of a discrete random variable X with probability distribution p(x) is given as
(The sum is over all values of x for which p(x)>0)We sometimes use the notation
E(X)=μfor this equivalence.
Definition 4.4
Note: Not all expected values exist, the sum above must converge absolutely, ∑|x|p(x)<∞.
Theorem 4.1 If X is a discrete random variable with probability p(x) and if g(x) is any real-valued function of X, then E(g(x))=∑g(x)p(x).
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Definitions 4.5 and 4.6
The variance of a random variable X with expected value μ is given by
V(X)=E[(X- μ)2]Sometimes we use the notation
σ2 = E[(X- μ)2]For this equivalence.
The standard deviation is a measure of variation that maintains the original units of measure.
The standard deviation of a random variable is the square root of the variance and is given by
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Theorem 4.2 For any random variable X and constants a and b.
1. E(aX + b) = aE(X) + b2. V(aX + b) = a2V(X)
Standardized random variable: If X has mean μ and standard deviation σ, then Y=(X – μ)/ σ
has E(Y)=0 and V(Y)=1, thus Y can be called the standardized random variable of X.
Theorem 4.3 If X is a random variable with mean μ, then V(X)= E(X2) – μ2
Tchebysheff’s Theorem. Let X be a random variable with mean μ and standard deviation σ. Then for any positive k,
P(|X – μ|/ σ < k) ≥ 1-1/k2
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Part 6. Bernoulli, Binomial and Geometric Distribution
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Let the probability of success is p, then the probability of failure is 1-p, the distribution of X is given by
p(x)=px(1-p)1-x, x=0 or 1
Where p(x) denotes the probability that X=x.
E(X) = ∑xp(x) = 0p(0)+1p(1)=0(1-p)+p= p E(X)=p
V(X)=E(X2)-E2(X)= ∑x2p(x) –p2=0(1-p)+1(p)-p2=p-p2=p(1-p) V(X)=p(1-p)
Bernoulli Distribution
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Binomial Distribution
Suppose we conduct n independent Bernoulli trials, each with a probability p of success. Let the random variable X be the number of successes in these n trials. The distribution of X is called binomial distribution.
Let Yi = 1 if ith trial is a success= 0 if ith trial is a failure,
Then X=∑ Yi denotes the number of the successes in the n independent trials.So X can be {0, 1, 2, 3, …, n}.
For example, when n=3, the probability of success is p, then what is the probability of X?
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Cont.
From the binomial formula, we can have
=
The mass function of binomial distribution:
A random variable X is a binomial distribution if1. The experiment consists of a fixed number n of identical trials.2. Each trial only have two possible outcomes, that is the Bernoulli
trials.3. The probability p is constant from trial to trial.4. The trials are independent.5. X is the number of successes in n trails.
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E(X)=np
Bernoulli random variables Y1, Y2, …, Yn, then
V(X)=np(1-p)
Bernoulli random variables Y1, Y2, …, Yn, then
V
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The geometric distribution function: P(X=x)=p(x)=(1-p)xp=qxp, x= 0, 1, 2, …., q=1-p
Geometric Distribution: Probability Function
P(X=x) = qxp = p[qx-1p] = qP(X=x-1) <P(X=x-1)
as q ≤ 1, for x=1, 2, …
A Geometric Distribution Function with p=0.5
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Geometric Series and CDF
The geometric series: {tx: x=0, 1, 2, …}
Sum of Geometric series: For |t|<1, we have =
Sum of partial series: =
Then we can verify
The cumulative distribution function:F(x)=P(X≤x)===1-qx+1
And P(X≥x)=1-F(x-1)=qx