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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 ?

PROBABILITY 535 Solution Let b stand for boy and g for girl. The sample space of the experiment is S = {(b, b), (g, b), (b, g), (g, g)} Let E and F denote the following events : E : ‘both the children are boys’ F : ‘at least one of the child is a boy’ Then E = {(b,b)} and F = {(b,b), (g,b), (b,g)} Now E ∩ F = {(b,b)} Thus Therefore P (F) = 1 3 and P (E ∩ F )= 4 4 1 P (E ∩ F) 4 1 = = P (E|F) = 3 3 P ( F) 4 Example 3 Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number? Solution Let A be the event ‘the number on the card drawn is even’ and B be the event ‘the number on the card drawn is greater than 3’. We have to find P(A|B). Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Then A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10} and A ∩ B = {4, 6, 8, 10} Also Then P(A) = 5 7 4 , P ( B) = and P (A ∩ B) = 10 10 10 4 P (A ∩ B) 10 4 = = P(A|B) = 7 7 P ( B) 10 Example 4 In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl? Solution Let E denote the event that a student chosen randomly studies in Class XII and F be the event that the randomly chosen student is a girl. We have to find P (E|F).

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