Probability Theory PDF

PROBABILITY THEORY Sample space: The sample space is the collection or totality of all possible outcomes of conceptual experiment. An event is a subset of the sample space. The class of all events associated with a given experiment is defined to be the event space. Let us describe the sample space...

PROBABILITY THEORY Sample space: The sample space is the collection or totality of all possible outcomes of conceptual experiment. An event is a subset of the sample space. The class of all events associated with a given experiment is defined to be the event space. Let us describe the sample space S, i.e. the set of all possible relevant outcomes of an experiments like tossing a single coin or rolling a die etc, e.g., S = {H, T } , S = {1, 2, 3, 4, 5, 6}. In both of these examples we have a finite sample space. Random Experiment : An experiment in which we know the set of all different possible outcomes and which is such that it is not possible to predict which one of the set will occur at any particular performance of the experiment is called a random experiment. It is denoted by E. Mutually Exclusive event: Two events A and B connected with a random experiment E, are said to be mutually exclusive if they cannot occur simultaneously. Symbolically, events A and B are mutually exclusive if A ∩ B = Φ or P(A∩ B ) =0. Classical Definition of Probability: Suppose the sample space S of a random experiment contains a finite ,say n(S) of even points. If m(A) is the number of event points of these n(S) event points are contained in the event A , connected with the 𝑚(𝐴) random experiment then the ratio is called the probability of occurrence of 𝑛(𝑆) the event A and is denoted by the symbol P(A). 𝑚(𝐴) Therefore, P(A) = 𝑛(𝑆) Axioms of probability: Axiom (i) : for any event A, P(A) ≥ 0 Axiom (ii) : for the certain event S, P(S) =1 Axiom (iii) : for a finite number of pair wise mutually exclusive events A1, A2 , … Of S , P(A1 U A2 U A3 U …) =P(A1) + P(A2) + P(A3) +….. Notations : Let S be the sample space of a random experiment E and A and B are two events connected with E, then i) P(A) denotes the probability of occurrence of A. ii) ̅̅̅ denotes the probability of non- occurrence of A P(Ac) or P( 𝐴) iii) P(AUB) or P(A+B) denotes the probability of occurrence of at least one of the events A and B. iv) P(A∩ B) or P(AB) denotes the probability of occurrence of both A and B. 𝐴 v) P( ) denotes the probability of occurrence of event A when it is known 𝐵 that B has already occurred. Properties of Probability 1. P (φ) = 0 2. P (A) ≤ 1 3. P (Ac) = 1 − P (A) 4. P (B ∩ Ac) = P (B) − P (B ∩ A) 5. 0≤ P(A) ≤ 1. Theorem of Addition: a) For any two events A and B P(A+B) = P(A) + P(B) – P(AB). If A and B are mutually exclusive then P(AB) = 0. b) For any three events A,B and C P(A+B+C) = P(A) + P(B) +P(C) – P(BC) – P(CA) – P(AB) + P(ABC) If A , B and C are mutually exclusive then P(A+B+C) = P(A) + P(B) +P(C). Conditional Probability: Let S be the sample space of a random experiment E and A and B be two events connected with E such that P(A) >0. Then the probability of occurrence of event B on the hypothesis that event A has actually occurred, is called conditional 𝐵 probability of B, given A and denoted by P( ). 𝐴 Theorem of compound probability: For two events A and B connected with a random experiment E ,the probability of their simultaneous occurrence is equal to the product of probability of A and the conditional probability of B on the hypothesis that the event A has actually occurred. Symbolically, 𝐵 P( AB) = P(A). P( ). 𝐴 Independent Event: We define two events A and B are independent if and only if P(AB) =P(A)P(B). Examples: Q) In a pack of 10 watches , 3 are known to be defective. If 2 watches are selected at random from the pack, what is the probability that at least one is defective? Sol: 2 watches from a pack of 10 watches can be selected at random in 10C2 = 45 ways. Let A denote the event that both the selected watches are non- defective. Now there are 7 non defective watches in the pack and 2 watches can be selected in 7C2 = 21 ways. Therefore by the classical definition of probability 21 7 P(A) = =. 45 15 Hence , the probability that at least one is defective = P( Ac) = 1 – P(A) 7 8 = 1-. = 15 15 Example 1. What is the chance that a leap year selected at random will contain 53 Sundays? Sol: We know that a leap year has 366 days. 366 days is composed of 52 complete weeks and 2 more days. So a leap year will contain 53 Sundays if the said two consecutive days contain one Sunday.  Two consecutive days of a week can be selected in the following different ways: ( Mon, Tue) ,( Tue, wed) ,(wed, Thu), ( Thu , Fri) , (Fri, Sat) ,(sat, Sun), (Sun, Mon). Let A denote the event that a leap year contains 53 Sundays. Clearly , A contains 2 equally likely event ,(sat, Sun), (Sun, Mon). 2 Required chance = P(A) = 7 3 5 3 𝐴 𝐵 Example 2. Given P(A) = ,P(B) = and P(AB) = Find P( ) and P( ).Are the two 8 8 4 𝐵 𝐴 events A and B are independent? Sol: For any two events A and B we have P(A+B) = P(A) + P(B) – P(AB). 3 5 3 1 Or, P(AB) = P(A) +P(B) – P(A +B)= + - = 8 8 4 4 𝐴 𝑃(𝐴𝐵) 2 𝐵 𝑃(𝐴𝐵) 2 Now, P( ) = = and P( ) = = 𝐵 𝑃(𝐵) 5 𝐴 𝑃(𝐴) 3 3 5 15 1 WE HAVE P(A).P(B) = x = and P(AB) =. 8 8 64 4 ∴ P(AB) ≠ P(A).P(B) ∴ Events A and B are not independent. Baye’s Theorem: An event X can occur if one of the mutually exclusive and exhaustive set of events A1 , A2, …,An occurs. Assume that the unconditional probabilities P(A1), P (A2), 𝑋 𝑋 𝑋..P(An) and the conditional probabilities P( ) , P( ) ,… P( )are given. Then the A1 A2 An conditional probability of an arbitrary event Ai on the hypothesis that the event X actually occurred is given by , 𝑋 𝐴𝑖 𝑃(𝐴𝑖 ) 𝑃( ) 𝐴𝑖 P( ) = 𝑋 𝑋 𝑋 𝑋 𝑃(𝐴1 ) 𝑃( ) +𝑃(𝐴2 ) 𝑃( )+⋯+𝑃(𝐴𝑛 ) 𝑃( ) 𝐴1 𝐴2 𝐴𝑛 Example 3: Three identical boxes contain red and white balls. The first box contains 3 red and 2 white, the second box 4 red and 5 white, and the third box 2 red and 4 white balls. A box is chosen at random and a ball is drawn from it. If the ball drawn is red, what is the probability that the second box is chosen? Sol: Let A1 , A2, A3 denotes the event of choosing the first, second and third box and X be the event of drawing red ball. clearly, 1 P(A1) =P(A2) = P(A3) = 3 𝑋 3 3 𝑋 4 4 𝑋 2 2 1 P( ) = = , P( ) = = , P( ) = = = 𝐴1 3+2 5 𝐴2 4+5 9 𝐴3 4+2 6 3 𝑋 𝑋 𝑋 P(X) = 𝑃(𝐴1 ) 𝑃 ( ) + 𝑃(𝐴2 ) 𝑃 ( ) + 𝑃(𝐴3 ) 𝑃 ( ) 𝐴 𝐴1 𝐴 2 3 1 3 1 4 1 1 62 =. +. +. =. 3 5 3 9 3 3 135 By Baye’s Theorem we get 𝑋 1 4 𝐴2 𝑃(𝐴2 ) 𝑃(𝐴 ). 10 2 3. 9 P( )= = 62 =. 𝑋 𝑃(𝑋) 31 135 Mathematical Expectation: Suppose a random experiment has n mutually exclusive and exhaustive outcomes corresponding to which a variable x takes the values x1 ,x2, …xnwith probabilities p1 ,p2, …pn Then the mathematical expectation of the variable x denoted by E(x) is defined as E(x) = p1x1 + p2x2 +…..+pnxn = ∑pixi If E(x) =m (Mean),the mathematical expectation of (x-m)2 is called variance of x. Var(x) = E(x-m)2.i.e. Var(x) =E(x2) – [E(x)]2. Example 4. A card is drawn from each of two well-shuffled packs of cards. Find the probability that at least one of them is an ace. Sol: Let us denote by A = event that the card from Pack I is an ace, B = event that the card from Pack II is an ace, It is required to find P(A + B). Since there are 4 aces in a pack of 52 cards, P(A) = 4/52 = 1/13. Similarly, P(B) = 1/13. The events A and B are independent, because the drawing of an ace or otherwise from does one pack not in any way affect the probability of drawing an ace from another pack. So, P(A + B) = P(A) + P(B) - P(AB) 1 1 1 PAB) = PA).P(B) = x = 13 13 169 Substituting the values, 1 1 1 25 P(A+B) = P(A) + P(B) –P(AB) = + - = 13 13 169 169.