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https://docs.google.com/document/d/1FJ5f1YTFCwZVwnZlzozwltUpS8HFL_Pt/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true probability and their applications; Conditional probability; Baye's theorem, Probability distribution of a ra ndom va riate; Binomial distributions and its properties Probability or chance is a measure of uncertainty in any event occurring. Life is full of uncertainties. Probability measures the degree of uncertainty between 0 and 1. Galileo was first to attempt probability while he was dealing with some problem relating to dice in gambling. The foundation of theory of probability was laid by B. Pascal and P. Fermat. Now probability is widely used in business and economics for making future predictions. BASIC CONCEPTS Experiment or trial An operation which results in some well defined outcomes. Random Experiment An experiment whose outcome cannot be predicted with certainty is called a random experiment. Tossing of a coin is a random experiment. Throwing a die is a random experiment. Sample Space The Set of all possible outcomes of an experiment is called the sample space of that experiment. It is denoted by S. When a coin is tossed sample space S = {H, T} When a die is thrown, sample space S = {1, 2, 3, 4, 5, 6} Event : A subset of the sample space is called an event. CHAPTER INCLUDES : Basic concepts Types of events Algebra of events Probability of an event Odds of an event Addition theorem Independent events Conditional probability Multiplication laws of probability Total probability theorem Baye's Theorem Probability distribution Binomial distribution Multinomial theorem Geometrical probability Solved examples Give the sample space when Three coins are tossed together. Two die are thrown together. Solution : The sample space is HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. The sample space shall consist of following 36 outcomes (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) TYPES OF EVENTS Simple event or elementary event An event is called a simple event if it is a singleton subset of the sample space S. For example, getting of three heads in simultaneous throw of 3 coins is a simple event. Mixed event or compound event A subset of the sample space S which contains more than one element is called a mixed event. For example, getting the same number on both die, when cast together is a compound event. Impossible event : Represented by empty set φ. For example, getting a number 7 when a die is thrown, is an impossible event. Sure event or certain event : The event represented by sample space S itself is a sure event or certain event. For example, the event of getting one red or black card, when a card is drawn from a well shuffled pack of cards is a sure event. Equally likely events : Events are said to be equally likely when one does not happen more often that the other. For example, the event of drawing a black card and the event of drawing a red card are equally likely, when a card is drawn from a pack of 52 cards. Mutually exclusive or disjoint events : A set of events is said to be mutually exclusive, if the happening of one excludes the happening of the others. Thus events A1, A2, , An are mutually exclusive if and only if Ai ∩ Aj = φ for all i ≠ j. For example, when husband and wife both appear for an interview for one post, the selection of husband and the selection of wife are mutually exclusive events. Exhaustive events (cases) : A set of events is said to be exhaustive if the performance of the experiment always results in the occurrence of at least one of them. For example, the events {1, 2}, {3, 4, 5} and {6} are exhaustive event for the sample space {1, 2, 3, 4, 5, 6}. If A1, A2,.........., An are exhaustive events then, A1 ∪ A2 ∪ ∪ An = S ALGEBRA OF EVENTS Let E and F be two events. Then E' or E or EC stands for the non-occurrence or negation of E. E ∪ F stands for the occurrence of at least one of E and F (Also denoted by E + F, E or F ) E ∩ F stands for the simultaneous occurrence of E and F. (Also denoted by EF or E and F ) E' ∩ F' stands for the non-occurrence of both E and F E ⊆ F stands for the occurrence of E implies occurrence of F. If E ∩ F = φ ⇒ E and F are mutually exclusive events. E–F denotes the occurrence of event E but not F E–F = E ∩ F' E ∪ F = F ∪ E and E ∩ F = F ∩ E (E ∪ F)' = E' ∩ F' and (E ∩ F)' = E' ∪ F' A die is cast. If the number on it is even, a coin is tossed and if odd, then two coins are tossed. List the sample space. Solution : Even numbers of die are (2, 4, 6) and odd numbers are (1, 3, 5). Hence S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, 1TT, 3HT, 3TH, 3TT, 3HH, 5HH,5HT, 5TH, 5TT} PROBABILITY Probability of occurrence of an event is a number lying between 0 and 1 i.e. 0 ≤ P(E) ≤ 1 Let S be the sample space, then the probability of occurrence of an event E is denoted by P(E) and is defined as P (E ) = n(E ) = number of elements in E n(S) number of elements in S = number of cases favourable to event E total number of cases If P(E) = 0 Event is impossible P(E) = 1 Event is sure P(E) 1 and P(E) 0 Remarks : More is the probability of an event, more are chances of its happening. 2. P(φ) = 0 & P(S) = 1 i.e. nothing outside sample space can occur. Find the probability of 5 Sundays falling in February month of a leap year. Solution : February in leap year has 29 days. 7 × 4 = 28, hence in First 28 days each weekday (Sunday, Monday..... Saturday) falls 4 times. The 29th day can be any of these seven week days. Hence S = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} and E = {Sunday} Hence P(E ) = 1 7 From a pack of 52 playing cards, a card is chosen at random. Find the probability of its being a court card a number card an honour card a number card smaller than 5 Solution : Court cards are Kings, Queens & Knaves. So no. of favourable cases are 3 × 4 = 12 Court cards are there among 52 cards. 12 C1 = 12 = 3 Hence probability of being a court card = 52 C1 52 13 There are 9 number cards (from 2, 3 10) in each suit. Hence a total of 9 × 4 = 36 number cards. 36 C1 = 36 = 9 Hence required probability = 52 C1 52 13 There are 4 × 4 = 16 honour cards (4 Aces, 4 Kings, 4 Queens and 4 Knaves). ∴ 16 C1 = 16 = 4 Required probability = 52 C1 52 13 Number cards smaller than 5 are 2, 3 and 4 in each suit, hence a total of 3 × 4 = 12 cards ∴ Required probability = 12 = 3 52 13 ODDS IN FAVOUR OR AGAINST If an event E can happen in a ways and fails to happen in b ways then; (i) P(E ) = a a + b (ii) P(Ec ) = b a + b The odds in favour of event E = a b The odds against event E = b a If P is the probability of an event E, then Probability of E not happening = 1 – P P Odds in favour of E = (1− P ) (1− P ) Odds against E = P A party thrown for families have boys, girls, ladies and gentlemen. When one member is chosen, odds against 27 being a boy is 5 7 ; odds in favour of it being a girl is 25 11 and probability of being a lady is 32 . Find the odds in favour of the chosen person being a gentleman. Solution : Let B is the event that a boy is chosen, G is the event of selecting a girl, L is the event of selection of a lady and M is the event of selection of a gentleman. All the four events are mutually exclusive and exhaustive hence P(B) + P(G) + P(L) + P(M) = 1 P(B) = 5 = 5 ; P(G) = 7 = 7 , P(L) = 11 hence, 5 + 27 32 7 + 25 32 32 P(M ) = 1− 5 − 7 − 11 = 1− 23 = 9 32 32 32 32 32 So odds in favour of M = 9 P(M ) = 32 = 9 1− P(M ) 1− 9 23 32 ADDITION THEOREM If p1, p2, .......... pn be the probabilities of n mutually exclusive events E1, E2, En, then the probability p, that any one of these events will happen is given by p = p1 + p2 + + pn or P(E1 ∪ E2 ∪ .......... ∪ En) = P(E1) + P(E2) + + P(En) If E and F are any two events (not mutually exclusive), then P(E ∪ F) = P(E + F) = P(E) + P(F) – P(E∩F) = P(E) + P(F) – P(EF) In general, If E1, E2, En are n events, then P(E1 ∪ E2 ∪ .......... ∪ En) = ∑P(Ei ) − ∑P(Ei ∩ E j ) + ∑P(Ei ∩ E j ∩ Ek ) – + (–1)n – 1 i =1 1≤i < j ≤n 1≤i < j 0 then. P(E1 Ι E2 Ι Ι En+1) = P(E1). P(E2 / E1).P(E3 / E1 Ι E2 ) P(En+1 / E1 Ι E2 Ι Ι En ) Special Cases : If E1, E2 En are independent events. Then P(E1 Ι E2 Ι En ) = P(E1 ) P (E2 )....... P(En ) The probability that none of E1,...... En happen is = (1 – P (E1)) . (1 – P(E2)). (1– P (En)) If p is the probability that an idependent event happens in one trial, then the probability that it will happen in a succession of k trials = pk Total Probability Rule : If {E }, i = 1, 2.......n be n events such that E Ι E = 0 for j and ≠ i exhaustive). Suppose P(Ei) > 0 (i = 1, 2 n). Then for any event A, P(A) = ∑P(Ei ) P (A / Ei ) = P(E1) P(A/E1) + + P(En) P(A/ En). i =1 Baye’s Rule : n Υ Ei i =1 S = S (they are mutually exclusive and Let {E } be mutually exclusive events such that P(E ) > 0 for i = 1, n and S = n Υ Ei . Let A be any event i i with P(A) > 0. Then for i = 1, 2, n P(Ei ).P(A / Ei ) i =1 P(Ei/A) = ∑P(Ei ).P(A / Ei ) i =1 Facts to Remember : Let E and F are mutually exclusive and exhaustive events and G be any other event. Then (i) G = (G Ι E) Υ (G Ι F) P(G) = P(E) P(G/E) + P(F) P(G/F) P(E ) ⋅P(G / E ) P(E/G) = P(E )⋅P(G / E ) + P(F )⋅P(G / F ) PROBABILITY DISTRIBUTION OF A RANDOM VARIATE A random variable which can take only finite or countably infinite number of values is called discrete random variable otherwise continuous random variable. If the values of a random variable together with the corresponding probabilities are given, it is called the probability distribution of random variable. e.g., the probability distribution of number of heads when two coins are tossed together is X : 0 1 2 P( X ) : 1 1 1 4 2 4 Binomial Distribution If n independent trials are performed and if p be the probability of success of each trial and q of its failures, then the probabilities of getting 0, 1, 2, 3, n successes are given by the respective terms of the Binomial expansion. (q+p)n = qn + nC qn–1 p + nC qn–2 p2 + + pn = P (0) + P(1) + P(2) + + P(n) The values of p and q remain the same for all trials in an experiment The probability of exactly r successes in n trials is given by P (r) = nC . qn–r . pr The probability of at least one success = P(1) + P(2) + ....... + P(n) = 1 – P(0) The probability of at least two successes = 1 – [P(0) + P(1)] and so on Note : The mean, variance and standard deviation of a binomial distribution are np, npq and respectively. A fair coin is tossed four times. Let X denotes the number of times a head is followed immediately by a tail. Work out the probability distribution of X. Also, find the mean and variance of X. Solution : The sample space associated with the random experiment is given by S = {TTTT, TTTH, TTHT, THTT, HTTT, TTHH, THTH, THHT, HTHT, HTTH, HHTT, THHH, HTHH, HHTH, HHHT, HHHH}. Now, {X = 0} = {TTTT, TTTH, TTHH, THHH, HHHH} {X = 1} = {TTHT, THTT, HTTT, THHT, THTH, HTTH, HTTH, HHTT, HTHH, HHTH, HHHT} {X = 2} = {HTHT} Thus P (X = 0) = 5 , P( X = 1) = 10 and P( X = 2) = 1 16 16 16 Therefore the probability distribution of X is given by sp. sp. x 0 1 2 P( X = x): 5 10 1 16 16 16 The mean of X = E( X ) = 0. 5 + 1. 10 + 2. 1 = 12 = 3 16 16 16 16 4 and E( X 2 ) = 02. 5 + 12. 10 + 22. 1 = 14 = 7 16 16 16 16 8 Therefore var ( X ) = E( X 2 ) − (E( X ))2 = 7 − 9 8 16 = 5 . 16 Use of Multinomial Theorem : Let a die having m faces (with numbers 1, 2, 3 m) be tossed n times. Then the probability that the sum of the numbers obtained on the die is p is given by the coefficient of xp in the expansion of (x + x 2 + x 3 + + xm )n mn SOLVED EXAMPLES Example 1: From a pack of 52 playing cards, find the probability of getting either a king or a queen or an ace. What is the probability of getting a king or a spade ? Solution : Let p1 = probability of getting a king p2 = probability of getting a queen and Then, p3 = probability of getting an ace p1 = p2 = p3 = 4 = 1 . 52 13 In first part of the question, since the three events are mutually exclusive therefore the probability of getting either a king or a queen or an ace = p1 + p2 + p3 (Addition Law) = 3 13 But in the second part the two events are not mutually exclusive, therefore the probability of getting a king or a spade is given by P(King) + P(Spade) – P(King ∩ Spade) = 4 + 13 − 1 52 52 52 = 16 = 4 . 52 13 Let the probabilities of solving a certain problem by three students A, B, C be p1 = 3 , p 4 2 = 2 , p 3 3 = 5 respectively. 6 Then find the probability that The problem will be solved by all the three students, The problem will not be solved by any of them and At least one of them will be able to solve the problem. Solution : All three events are independent, therefore using Multiplication Law, the probability is given by p × p × p = 3 × 2 × 5 = 5 1 2 3 4 3 6 12 Let q1, q2, q3 be the probability that problem will not be solved by A, B, C respectively, then q1 = 1− p1 = 1 , q 4 2 = 1− p2 = 1 ,q 3 3 = 1− p3 = 1 . 6 Therefore the required probability = q1 q2 × q3 = 72 The probability that at least one of them will be able to solve the problem = 1− q q q = 1− 1 = 71 . 1 2 3 72 72 Example 3 : If m different toys are distributed among n kids, find the probability that a particular kid will get exactly k (