ME 233 Advanced Control Il Lecture 2 Introduction to Probability Theory (ME233 Class Notes pp. PR1-PR3)
. Sample Space and Events • Probability function • Discrete Random Variables • Probability mass function, expectation and variance
Assume: We do an experiment many times. • – Each time we do an experiment we call that a trial . The outcome of the experiment may be different at each trial.した。 The ith possible outcome of the experiment
Sample Space   Ω : The space which contains all possible outcomes of an experiment.
W i : The ith possible outcome of the experiment Each outcome is an element of
Experiment: A situation whose outcome  depends on chance - throwing a die once Sample Space Ω The set of all possible outcomes of an experiment 25 = {0,0,0,0,0,0,0 Events Event  Sj : Is a subset of the union of the sample space  Ω and the empty set   φ If a sample space has  n  outcomes:
There are   2  n   events:
Experiment: throwing a die once Ω = {□, □, □, □, □, □, □ Outcomes: elements of the sample space S Events: Are subsets of the sample space S An event occurs if any of the outcomes in that event occurs. Empty subsets are null or impossible events 7
Experiment: throwing a die once
Some events: •    The event E of observing an even number of dots: E =  {ロ:ロ:ロ:ロ:ロ} The event  O of observing an odd number of dots: Example: throwing a pair of dice (one red and one blue) the sample space has 36 outcomes: The event L of obtaining the number 7 is L = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}
L occurs if any of the outcomes in L occurs.

For a sample space  Ω = {ω1, ω2,۰.۰, wn} And the set of all events  S = {S'1, ... , S'2n} Union of two events (or): Si U Si = {wm | wm E Si or wm E Si} Intersection of two events (and): Si ∩ S j ＝ {wm | wm ∈ S j and wm ∈ S j } Complement of an event (not): \Si = {wm | wm ∈ Ω ∪ φ and wm ∈ Si}
. Union of two events: Si U Si Intersection  of two events: Si N Si Complement of an event:
•  The complement of an event A , denoted by A c , is the set of outcomes that are not in A •  Ac   occurring means that A does not occur
•  The intersection of two events  A and B , denoted by A ∩ B, is the set of outcomes that are in A, and B. •  If the event A ∩ B occurs, then both A and B OCCur •  Events A and B are mutually exclusive_if they cannot both occur at the same time, i.e. if A N B = 0 Example of Intersection of two events Experiment: throwing of a dice once Ω = {□,□,□,□,□,□,□} •    Events E and O are mutually exclusive
• The union of two events A and B, denoted by A U B, is the set of outcomes that are in A, or B, or both •   If the event A U B occurs, then either A or B or both occur
We now consider the probability that a certain event occurs. Recall: An event occurs if any of the outcomes in that event occurs. The probability of event A will be denoted by P( A )
A number between 0 and 1, inclusive, that indicates how likely an event is to occur. – An event with probability of 0 is a null event. – An event with probability of 1 is a certain event. . Probability of event A is denoted as P(A). •  The closer P(A) to 1, the more likely is A to happen. 17
The probability of event A is Possible outcomes associated with A P(A) = Total possible outcomes (Assumes each outcome is equally likely)
Assigning Probability - Frequentist  approach •  An experiment is repeated N times under essentially identical conditions •  if the event A occurs m times and n is large
n 19 Dice example Experiment: throwing a fair die once Ω = {D,D,D,B,B,B,B}}  Ω = {1,2,3,4,5,6} •  P(Ω) = 1 •  P(1) = 1/6,   P(3) =1/6, P(6) = 1/6 •  P(even number) = 3/6 = 1/2 •  P(odd number) = 3/6 = 1/2
Example: In poker you are dealt 5 cards from a deck of 52 . What is the probability of  being dealt four of a kind? e.g. 4 aces or  four kings, and so fourth? P(four of a kind) =?
Solution: 1. There are only 48 possible hands containing  4 aces, another 48 containing 4 kings, etc. 2. 13 x 48  possible “four of a kind” hands. Thus, there are 3. The possible number of hands is obtained from the combination formula for “52 things taken 5 at a time”:
Probability Space The probability space is the triple: (Ω, S, P) Where •  is the sample space . S the set of all possible events P : S → [0, 1]   is the probability function Probability function Probability function:    P : S → [0, 1] Satisfies 3 axioms: 1.
AS; E S 2. 3. P(S i ∪ S i ) = P(S i ) + P(S i ) if S i ∩ S i = ∅ where Si, Si E S 24
•  The complement of an event A , denoted by A c , is the set of outcomes that are not in A •  Ac occurring means that A does not occur P(A c ) = 1 – P(A)
•  Two events are independent if P(A ∩ B) = P(A) × P(B) •   Intuitively, two events are independent if the events do not influence each other: – Event A occurring does not affect the chances of B occurring, and vice versa.
Experiment: throwing a pair of dice (one red and one blue) 36 possible outcomes The probability of throwing a red 1 and a blue 5 is P(1 N 5 ) = 1/36
•  Recall: If A and B are mutually exclusive P(A U B ) = P(A) + P(B) • If A and B are not mutually exclusive P(A U B ) = P(A) + P(B) - P(A ∩ B )
•  If A and B are not mutually exclusive P(A U B ) = P(A) +  P(B) - P(A NB ) don’t count twice! Example Experiment: throwing a pair of dice (one red and one blue) •  P(L) = the probability of obtaining  a 7 L = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} P(L) = P(1,6) + P(2,5) + P(3,4) + P(4,3) + P(5,2) + P(6,1) P(L) = 6 / 36 = 1 / 6 Joint Probability Let A and B be two events
is often called the joint probability of A and B
are often called the marginal probabilities of A and  B Conditional Probability Let A and B be two events and    P(B) ≠ 0 The conditional probability of event A given that event B has occurred is P(A|B) P(A n B) P(B) Bayes’ Rule Let A and B be two events P(A|B)P(B) = P(B|A)P(A) P(An B) ll Independence The following are equivalent: 1.  A and B  are independent
Array of Probabilities Let C and D be two chance experiments. Set of disjoint events associated with C
Set of disjoint events associated with D
We can construct:
There are 4 balls in one jar, 2 balls are red and two balls are black. •  A person can remove a ball from the jar two times, without seeing the balls inside the jar.
What is the probability of removing a red ball after having removed a red ball the first time? To answer this question, lets build the table of probabilities.
What is the probability of removing a red ball after having removed a red ball the first time? To answer this question, lets build the table of probabilities. Labels: Example: Probability of picking R 1 the first time?
Probability of picking R 2 with only 3 balls left?
(second time) R 1
Probability of picking R 1 the first time and R 2 the second time? P(R1 ∩ R2) = 1/4 × 1/3 = 1/12
Probability of picking red balls consecutively Probability of event A: picking a red ball the first time and a red ball the second time? •   Event B: Picking R 1 first and R 2 second •   Event C: Picking R 2 first and R 1 second Mutually exclusive events P(A) = P(BUC)
What is the probability of picking a red ball the second time after having picked a red ball the first time?
Given a sample Space Ω, a random variable X is a function that assigns to each outcome a unique numerical value. . Example: throwing of a die once Ω =    {𝐄, 𝐄, 𝐄, 𝐄, 𝐄, 𝐄} 3.4.5
Example:_throwing of a die once
•  In this case, the random variable  X only takes discrete values
•  The discrete random variable  X is defined by the probability mass function
the probability that, after throwing a die, X will be equal to x i . Discrete random variable For a fair die, the probability mass function of the random variable  X is P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1 / 6 the probability mass function satisfies: 6 1 Expected value •  For a discrete random variable  X taking on the N possible values
the expected value or mean of X is defined by
Expected value of a function •  For a discrete random variable  X taking on the N possible values
and the real-valued function f the expected value or mean of Y=f(X) is defined by
Example: For a fair dice,
the expected value or mean of X
52 Variance and standard deviation •   For a discrete random variable  X taking on the N possible values
the variance of  X is defined by N
where   Ơχ is the standard deviation of X 53
the variance and standard deviation of X are
54 Cumulative Distribution Function •   The cumulative distribution function (CDF) for a discrete random variable X is
Find index R such that
• The cumulative distribution function (CDF) for a discrete random variable X is k F x Xk Xk+1 j=1 Sum of two uniform independent random variables • Let X  and  Y be 2 independent  random variables with constant probability mass function Z = X + Y • Let •  The probability mass function of Z  will not be constant
Experiment: throwing a pair of fair dice (red and blue) the sample space has 36 outcomes: each outcome has a  1/36 probability of occurring
. Define the random variable Z associated with the event of observing the total number of dots on both dice after each throw Z = k when the throw results in the number k Throwing two fair dice number of outcomes 36 Z  only takes discrete values
(2 = Throwing two fair dice probability of each outcome 1/36 we now estimate: Z=2 →   P(2) = 1 / 36 Z=7 →  P(7) = 6 / 36 Z=12 →   P(12) = 1 / 36 Z= 3 →   P(3) = 2 / 36 P(4)=3/36 Throwing two fair dice
The probability mass function is
Ω = the probability mass function satisfies: