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Probability

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Type: NoteCourse: Placement Preparation Specialization: Quantitative AptitudeOffline Downloads: 2Views: 66Uploaded: 2 months ago

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PROBABILITY Chapter 531 13 PROBABILITY ™ The theory of probabilities is simply the Science of logic quantitatively treated. – C.S. PEIRCE ™ 13.1 Introduction In earlier Classes, we have studied the probability as a measure of uncertainty of events in a random experiment. We discussed the axiomatic approach formulated by Russian Mathematician, A.N. Kolmogorov (1903-1987) and treated probability as a function of outcomes of the experiment. We have also established equivalence between the axiomatic theory and the classical theory of probability in case of equally likely outcomes. On the basis of this relationship, we obtained probabilities of events associated with discrete sample spaces. We have also studied the addition rule of probability. In this chapter, we shall discuss the important concept of conditional probability of an event given that another event has occurred, which will be helpful in understanding the Bayes' theorem, multiplication rule of Pierre de Fermat probability and independence of events. We shall also learn (1601-1665) an important concept of random variable and its probability distribution and also the mean and variance of a probability distribution. In the last section of the chapter, we shall study an important discrete probability distribution called Binomial distribution. Throughout this chapter, we shall take up the experiments having equally likely outcomes, unless stated otherwise. 13.2 Conditional Probability Uptill now in probability, we have discussed the methods of finding the probability of events. If we have two events from the same sample space, does the information about the occurrence of one of the events affect the probability of the other event? Let us try to answer this question by taking up a random experiment in which the outcomes are equally likely to occur. Consider the experiment of tossing three fair coins. The sample space of the experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
532 MATHEMATICS 1 to each sample point. Let 8 E be the event ‘at least two heads appear’ and F be the event ‘first coin shows tail’. Then E = {HHH, HHT, HTH, THH} and F = {THH, THT, TTH, TTT} Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH}) Since the coins are fair, we can assign the probability 1 1 1 1 1 + + + = (Why ?) 8 8 8 8 2 P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT}) = and 1 1 1 1 1 + + + = 8 8 8 8 2 E ∩ F = {THH} = Also 1 8 Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is the probability of occurrence of E? With the information of occurrence of F, we are sure that the cases in which first coin does not result into a tail should not be considered while finding the probability of E. This information reduces our sample space from the set S to its subset F for the event E. In other words, the additional information really amounts to telling us that the situation may be considered as being that of a new random experiment for which the sample space consists of all those outcomes only which are favourable to the occurrence of the event F. Now, the sample point of F which is favourable to event E is THH. with P(E ∩ F) = P({THH}) = Thus, Probability of E considering F as the sample space = 1 , 4 1 4 This probability of the event E is called the conditional probability of E given that F has already occurred, and is denoted by P (E|F). or Probability of E given that the event F has occurred = 1 4 Note that the elements of F which favour the event E are the common elements of E and F, i.e. the sample points of E ∩ F. Thus P(E|F) =
PROBABILITY 533 Thus, we can also write the conditional probability of E given that F has occurred as P(E|F) = = Number of elementary events favourable to E ∩ F Number of elementary events which arefavourable to F n (E ∩ F) n (F) Dividing the numerator and the denominator by total number of elementary events of the sample space, we see that P(E|F) can also be written as n(E ∩ F) P(E ∩ F) n(S) = P(E|F) = n(F) P(F) n(S) ... (1) Note that (1) is valid only when P(F) ≠ 0 i.e., F ≠ φ (Why?) Thus, we can define the conditional probability as follows : Definition 1 If E and F are two events associated with the same sample space of a random experiment, the conditional probability of the event E given that F has occurred, i.e. P (E|F) is given by P(E|F) = P (E ∩ F) provided P(F) ≠ 0 P (F) 13.2.1 Properties of conditional probability Let E and F be events of a sample space S of an experiment, then we have Property 1 P (S|F) = P(F|F) = 1 We know that Also P (S|F) = P (S ∩ F) P (F) = =1 P (F) P (F) P(F|F) = P (F ∩ F) P (F) = =1 P (F) P (F) Thus P(S|F) = P(F|F) = 1 Property 2 If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then P((A ∪ B)|F) = P (A|F) + P(B|F) – P ((A ∩ B)|F)
534 MATHEMATICS In particular, if A and B are disjoint events, then P((A∪B)|F) = P(A|F) + P(B|F) We have P((A ∪ B)|F) = = P[(A ∪ B) ∩ F] P (F) P[(A ∩ F) ∪ (B ∩ F)] P (F) (by distributive law of union of sets over intersection) = P (A ∩ F) + P (B ∩ F) – P (A ∩ B ∩ F) P (F) = P (A ∩ F) P (B ∩ F) P[(A ∩ B) ∩ F] + − P(F) P(F) P(F) = P (A|F) + P (B|F) – P ((A ∩ B)|F) When A and B are disjoint events, then P ((A ∩ B)|F) = 0 ⇒ P ((A ∪ B)|F) = P (A|F) + P (B|F) Property 3 P (E′|F) = 1 − P (E|F) From Property 1, we know that P (S|F) = 1 ⇒ P (E ∪ E′|F) = 1 since S = E ∪ E′ ⇒ P (E|F) + P (E′|F) = 1 since E and E′ are disjoint events Thus, P (E′|F) = 1 − P (E|F) Let us now take up some examples. Example 1 If P (A) = 7 9 4 , P (B) = and P (A ∩ B) = , evaluate P (A|B). 13 13 13 4 P (A ∩ B) 13 4 Solution We have P (A|B) = = = 9 9 P ( B) 13 Example 2 A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ?

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