University of California, Los AngelesDepartment of StatisticsStatistics 100A Instructor: Nicolas ChristouProbabilityProbability: A measure of the chance that something will occur.1. Random experiment:A process that results in one of possible outcomes. The outcomescannot be predicted with certainty.Examples: Flip a coin, roll a die, roll two dice, draw a card, etc.2. Sample space of a random experiment:It is the list of all possible outcomes of the random experiment,denoted with S.Examples:a. Flip a coin: S = {H, T }.b. Flip two coins: S = {HH, T T, HT, T H}.c. Roll a die: S = {1, 2, 3, 4, 5, 6}.d. Draw a card: S = {A♣, A♠, A♦, A♥, · · ·} (all 52 cards).3. Event:It is the outcome of an experiment, denoted with uppercase let-ter. It is a subset of the sample space.Examples:a. Flip a coin: A = {H}.b. Roll a die: A = {even number}, B = {odd number}, C = {1, 2, 3}.d. Draw a card: A = {Ace}, C = {Clubs}.4. Probability of an event A for equally likely outcomes:P (A) =number of ways in which A occursnumber of ways in which all outcomes occur1Examples:a. Draw a card. Let A = {Ace}. Then P (A) =452.b. Roll two dice. There are 6 × 6 possible outcomes. The sum of the two numbers rolledare shown below:1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 94 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12Let A = {sum=5}. Then P (A) =436.4. Basic principle of counting: If an experiment has m outcomesand if for every outcome of this experiment there are n out-comes of another experiment then all together there are m × noutcomes.Examples:a. Roll two dice: 6 × 6 = 36 outcomes.b. Flip two coins: 2 × 2 = 4 outcomes.Similarly:a. Roll three dice: 6 × 6 × 6 = 216 outcomes.b. Flip three coins: 2 × 2 × 2 = 8 outcomes.2Union of two events A, BThe union of two events A, B, denoted A ∪ B, is a new event.It is defined as the event containing all outcomes in A or B orBOTH. Similarly the union of n events A1, A2, · · · , Anis denotedwith A1∪ A2∪ · · · ∪ An. Key word: OR.Example: Suppose 50 students can be classified by their major and year as follows:Sop. Jun. Sen. TotalEcon 10 20 6 36Math 5 4 5 14Total 15 24 11 50A student is selected at random from this group of 50 students. Let A = {student is senior}, andB = {student is math major}. Find P (A ∪ B).Intersection of two events A, BThe intersection of two events A, B, denoted A ∩ B, is a new event.It is defined as the event containing all outcomes that belong to bothA and B. Similarly the intersection of n events A1, A2, · · · , Anisdenoted with A1∩ A2∩ · · · ∩ An. Key word: AND.Example: Using the table above find P (A ∩ B).Example: Draw a card at random. Let A = {card is clubs} and B = {card is ace or 4}. FindP (A ∩ B). Also find P (A ∪ B). Use a Venn diagram to show these events.3Algebra of setsa. Cumulative laws:A ∪ B = B ∪ A.A ∩ B = B ∩ A.b. Associative laws:A ∪ B ∪ C = (A ∪ B) ∪ C = A ∪ (B ∪ C).A ∩ B ∩ C = (A ∩ B) ∩ C = A ∩ (B ∩ C).c. Distributive laws:(A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)(A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)Mutually exclusive eventsIt is said that two events A, B are mutually exclusive, (or disjoint) ifA, B have no outcomes in common. It follows that A, B are disjointif A ∩ B = {∅}. Therefore, if A, B are disjoint then P (A ∩ B) = 0.Example: Roll a die. Let A = {even number}, B = {odd number}. The events A, B are mutuallyexclusive and we can represent them as follows using a Venn diagram.4Axioms of Probability1. 0 ≤ P (A) ≤ 1.2. P (S) = 1.3. The union for any sequence of mutually exclusive events A1, A2, · · ·is equal to P (A1∪ A2· · ·) = P (A1) + P (A2) + · · ·Let A, B two mutually exclusive event. Then the union of thetwo events is equal to P (A ∪ B) = P (A) + P (B).Complement eventsThe complement of an event A, denoted with A0or Ac, is defined tobe the event that contains all the outcomes in the sample space thatdo not belong to A.Are A, A0mutually exclusive?5Conditional ProbabilityLet two events A and B. Suppose that event B has already occured.Given this information, what is the probability of event A. This iscalled “conditional probability” denoted with P (A|B) and we read:the probability of A given B. It is equal to:P (A|B) =P (A ∩ B)P (B)Similarly,P (B|A) =P (A ∩ B)P (A)Example: Roll a die. Let A = {2}, and B = {even number}. Draw the two events using a venndiagram. Find P (A|B).6• Multiplication ruleLet events A, B. The probability of A and B (i.e. the probabilityof the intersection A, B) can be computed as follows:P (A ∩ B) = P (A|B)P (B) = P (B|A)P (A).Example:Draw two cards without replacement. Let’s define the events A1, A2as follows:A1= {the first card is an ace}, and A2= {the second card is an ace}. Find the probabilitythat both cards are aces.• Independent eventsThe events A, B are independent if the occurrence of one ofthem does not affect the occurrence of the other event. If A, Bare independent then, P (A|B) = P (A), and the multiplicationrule is: P (A ∩ B) = P (A)P (B).Example: Draw two cards with replacement. Let’s define the events A1, A2as follows:A1= {the first card is an ace}, and A2= {the second card is an ace}. Find the probabilitythat both cards are aces.7• Addition ruleLet events A, B. The probability of A or B or BOTH (i.e. theprobability of the union A, B) can be computed as follows:P (A ∪ B) = P (A) + P (B) − P (A ∩ B). It follows directly fromthe Venn diagram.• If A, B are mutually exclusive events then P (A ∩ B) = 0, andthe addition rule is:P (A ∪ B) = P (A) + P (B) (this is Axiom 3).• If A, B are independent, can they be mutually exclusive?8Examples:1. Draw two cards with replacement. Let’s define the events A1, A2as follows:A1= {the first card is an ace}, and A2= {the second card is an ace}. Find the probabilitythat the first card is an ace or the second card is an ace, or both cards are aces. Or simplywe can ask: Find the probability that we observe at least one ace.2. Two dice are rolled 10 times. Find the probability that in these 10 rolls we observe the sumof 5 at least once.9De Morgan’s LawVery useful!Using De Morgan’s law we can find …