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**Chapter Two Probability**

STAT 111 Chapter Two Probability

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Probability Many statistical principles and procedures are based on the important concept of probability. The purpose of this chapter is to define such concept and discuss some of its properties. A random experiment is an experiment in which All possible outcomes of the experiment are known in advance, Any performance of the experiment results in an outcome that is not known in advance, The experiment can be repeated under identical conditions

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**Examples Tossing a coin once or several times.**

Obtaining blood types from a group of individuals. Determining the sex of a newborn. Tossing two dice.

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Definition Definition: The sample space of an experiment, denoted by S, is the set of all possible outcomes of that experiment. Sample spaces are either Discrete : which contains a finite number of elements , or and infinite but countable number of elements . Continuous : which contains an infinite number of sample points constituting a continuum , such as all points on a line segment or all the points in a plane . S Discrete Continuous (infinite) Infinite (Countable) Finite

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**S = {HH, HT, TH, TT} n(S) = 4, discrete**

Example Describe an appropriate sample space for the experiments below. Determine the number of elements and state whether the sample space is discrete or not. A coin is tossed two times. طرق حساب فراغ العينة إذا كان محدود أولاً باستخدام الشجرة البيانية H T S = {HH, HT, TH, TT} n(S) = 4, discrete

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**Example (continued) A coin is tossed and a dye is rolled**

ثانياً باسخدام الجداء الديكارتي S = (H,T)x(1,2, ...,6) S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}, n(s) = 2x6=12, discrete Two numbers are selected from the set {1, 2, 3} without repetition of digits. ثالثاً باستخدام الشبكة التربيعية S = {(1,2), (1,3), (2,1), (2,3), (3,1), (3,3) } n(S) = 3x2=6, discrete (1,2) 3 2 1 (1,3) (3,2) (2,1) (3,1) (2,3)

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**Example (continued) A coin is tossed until the first head appears.**

S = {H, TH, TTH, ...,∞), and so the coin is tossed an infinite number of times, here ∞, refers to the case when a head never appears. A light bulb is observed so that the length of its useful life might be recorded. S = {(0, ∞)}, since one could not say with certainty that the bulb would have burned out by any given time => S is continuous.

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Events Definition An event is any subset of the sample space S. An event A occurs, if the outcome of the experiment is in A. Example Two cards are drawn, randomly with replacement from three cards carrying the number 1,3, 5. Describe the sample space, find the events A= {sum of the two numbers is 5 } B= {sum of the numbers is at least 6} A=Φ B= {(1,5), (3, 3), (3, 5), (5,1), (5, 3), (5, 5)} (1,3) 5 3 1 (1,5) (5,3) (3,1) (5,1) (3,5) (3,3) (5,5) (1,1)

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Example Example: An experiment consists of rolling a die until a 3 appears, Construct a sample space for this experiment. Let 3 denotes 3 appears , let A denote 3 does not appear S = { 3 , A 3 , AA3 , …} . List the elements in E , the event that 3 appears before the fifth roll E = { 3 , A3 , AA3 , AAA3 }

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**Some Relations from Set Theory**

An event is nothing but a set, so that relationships and results from elementary set theory can be used to study events. The following concepts from set theory will be used to construct new events from given events. The union of two events A and B denoted by A B and read 'A or B' is the event consisting of all outcomes which are either in A or in B or in both events, that is A U B= {x ϵS :x ϵA or xϵ B} If A1, A2, ... are events, the union of these events, denoted, by is defined to be that event which consists of all points that are in An for at least one value of n = 1,2, ...

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**Some Relations from Set Theory**

The intersection of two events A and B, denoted by A B and read 'A and B' is the event consisting of all outcomes which are in both A and B, that is, A ∩ B={x ϵ S :x ϵ A and x ϵ B} If A1, A2, ... are events, the intersection of these events, denoted by , is defined to be that event consisting of all points that are in all of An, n= 1 , 2 , … The complement of an event A , denoted by A ( or Ac ) is the set of all outcomes in S which are not contained in A . That is Ac = { x ϵ S : x ϵ A }

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**Some Relations from Set Theory**

Definition When A and B have no outcomes in common (the intersection of A and B contains no outcomes, i.e. A ∩ B= Φ), they are said to be mutually exclusive of disjoint events. Definition A nonempty collection of subsets F of a set S is called a σ-filed of subsets of S provided the following properties hold.

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**Probability A Definition**

A probability measure P (or simply probability) is a real-valued function having domain F satisfying the following properties. Axioms of Probability: 1- 0≤ P(A) ≤ ϵ F 2- P(S)=1 3- If An, n = 1,2, ..., are mutually disjoint sets in S, then A probability space , denoted by ( S , F , P ) is a σ– field of subsets F and a probability measure P defined on S . A

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**Probability Definition**

An assignment of probability is said to be equally likely (or uniform) if each elementary event in S is assigned the same probability. Thus if S contains n elements wi, i.e. if S = { w1, …,wn} then

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**Properties of probability**

In the following, some additional properties of probability measure P will be presented. These properties follow from the definition of a probability measure. P( Φ ) = 0 Let A1, A2, ..., An a collection of pairwise disjoint events in S, then If A is an event in S, then P(A)=1-P(A) Let A and B be events in S , then P (A ) = P (A ∩ B ) + P ( A ∩ Bc) Let A and B be event in S, then P(A U B ) = P(A)+ P(B) - P(A ∩ B) P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B∩C) If A B , then P ( A ) ≤ ( B )

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Example 1 Example 1: If two dice are rolled what is the probability that The sum of upturned faces will equal 7? The sum of upturned faces will equal 2 or 12? The sum of upturned faces will be an even number or a number less than 6? Let A = {sum is even), B = {number less than 6} 6 (1,6) (2,6) (3, 6) (4, 6) (5,6) (6, 6) 5 (1,5) (2,5) (3, 5) (4,5) (5, 5) (6,5) 4 (1,4) (2,4) (3, 4) (4, 4) (5, 4) (6, 4) 3 (1,3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) 2 (1,2) (2,2) (3, 2) (4, 2) (5, 2) (6, 2) 1 (1,1) (2, 1) (3,1) (4,1) (5, 1) (6,1)

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**Example 2 Given P(A) = 0.59, P(B) = 0.30, P (A ∩ B) = 0.21, find**

A Ac Total B 0.21 0.09 0.3 Bc 0.38 0.32 0.7 0.59 0.41 1 Given P(A) = 0.59, P(B) = 0.30, P (A ∩ B) = 0.21, find P ( A U B ) = – 0.21 = 0.68 P (A ∩ Bc) = P ( A ) – P ( A ∩ B ) = = 0.38 P ( Ac U Bc ) = 1- P ( A ∩ B ) = 1 – 0.21 = 0.79 or P (AcUBc)=P(Ac)+P(Bc)–P(Ac∩Bc)= =0.79 P ( Ac ∩ Bc) = 1- P (A U B ) = =0.32

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Example 3 If A and B are two events such that A B. What is P(A B), what is P(A ∩ B) and what is P(A ∩ Bc)? P(A B)=P(B) P(A ∩ B)=P(A) P(A ∩ Bc)=P(A)- P(A ∩ B) = P(A)- P(A)=0 S B A

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**Example 4 If A, B and C are mutually exclusive events and P(A) = 0.2,**

P(B) = 0.3, and P(C) = 0.2, find P(A U B U C)? P(A U B U C)= P(A) + P(B) + P(C) = = 0.7 P(Ac ∩ (B U C))? P(Ac ∩ (B U C))=P(B) + P(C) = = 0.5 0.3 0.2 A B C S A B C S

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Example 5 A two-digit number is formed by randomly selecting, with replacement, digits from the set {6, 7, 8, 9}. Find the probability; The two digits are the same? n(S)=4x4=16 n(two digits are the same)=4x1=4 P(two digits are the same)=4/16=1/4 The number is odd? n(number is odd)=4x2=8 P(number is odd)=8/16=1/2 Remember

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Remember Example 6 If 10 people, including A and B, are randomly arranged in a line, What is the probability that A and B are next to each other? What would probability be if the people were randomly arranged in a circle?

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Independent Events Suppose that two events A and B occur independently of one another in the sense that the occurrence or nonoccurrence of either of them has no relation to the occurrence or nonoccurrence of the other, that is the probability that A and B will occur is equal to the product of their individual probabilities. Definition Two events A and B independent if and only if P ( A ∩ B ) = P ( A ) P ( B ) More generally , for n ≥ 3 , events A1 , A2 , … , An are independent if P ( A1 ∩ A2 ∩ … ∩An) = P (A1) P(A2)P(A3) …P ( An) And if any subcollection containing at least two but fewer than n events are independent . Events that are not independent are said to be dependent. Note that independence of events is not to be confused with disjoint or mutually exclusive events. Result Assuming that A and B are independent events, then the events Ac and B, A and Bc, Ac and Bc are also independent.

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Example 1 Let two fair coins be tossed, let A = {head on the second throw}, B = {head on the first throw}. Show that A and B are independent. S = {HH, HT, TH, TT} A = {HH, TH} B = {HH, HT} A∩B={HH} P(A∩B)=1/4 P(A)P(B)=½ x ½=1/4= P(AB) A and B are independent

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Example 2 Two students A and B are both registered for a certain course. If student A attends class 80 percent of the time and student B attends class 60 percent of the time, and if the absence of the two students are independent, what is the probability that at least one of the two students will be in class on a given day? P(A) = 0.8, P(B) = 0.6 Ac and Bc are independent A and B are also independent P ( A U B ) = P ( A ) + P (B) – P (A ) P (B ) = – 0.6x0.8=0.92

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Example 3 A coin is biased so that a head is twice as likely to occur as a tail, If the coin is tossed 3 times, what is the probability of getting 2 tails and 1 head? S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, let P(H) = 2W, P(T) = 1W 2W + 1W = W=1/3 P(H) = 2/3 and P(T) = 1/3 A = {TTH,THT,HTT}, P(TTH) = P(T) P(T) P(H)= 1/3 x 1/3 x 2/3 = 2/27 by independence P(THT) = P(T) P(H) P(T)= 1/3 x 2/3 x 1/3 = 2/27 by independence P(HTT) = P(H) P(T) P(T)= 2/3 x 1/3 x 1/3 = 2/27 by independence P(A) = P(TTH) + P(THT) + P(HTT)= 2/27 + 2/27 + 2/27=6/27=2/9

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**Example 4 P ( at least one lands head ) = P (A1 U A2 U A3)**

Suppose that three balanced coins are tossed. Find the probability that at least one of them lands head. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, A1=first coin lands head= {HHH, HHT, HTH, HTT} A2=second coin lands head={HHH,HHT,THH,THT} and A3=third coin lands head={HHH, HTH, THH, TTH} P(A1)=4/8=1/2 P(A2)=4/8=1/2 P(A3)=4/8=1/2 P ( at least one lands head ) = P (A1 U A2 U A3) Remember

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Example 5 A system containing five components is arranged in the manner shown in Figure - 1, where the probabilities given indicate the chance that the component will work. If we assume that whether a component works or not is independent of whether any other component is working or not. what is the probability that the system will work? P(system work) =P(A ∩ (B U C) ∩ (D U E)) = P(A)P(B U C)P(D U E) since independent P(B U C)=1- P(Bc ∩ Cc)= 1- P(Bc)P( Cc)=0.995 since independent Or ( B U C ) = P ( B ) + P ( C ) – P (B ) P (C) =0.995 since independent Similarly, P(D U E)= 1- P(Dc ∩ Ec)= 1- P(Dc)P( Ec)=0.9979 P(system work) = P(A)P(B U C)P(D U E)=0.98x0.995x0.9979=0.973 P(E)=0.97 P(D)=0.93 P(C)=0.95 P(B)=0.90 P(A)=0.98

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