Let A,B,C be three events. If the probability of occurring exactly one event out of A and B is $$1$$−a, out of B and C is $$1$$−$$2a$$, out of C and A is $$1$$−a and that of occurring three events simultaneously is $$a2$$, then the probability that at least one out of A,B,C will occur is

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Let A,B,C be three events. If the probability of occurring exactly one event out of A and B is $$1$$−a, out of B and C is $$1$$−$$2a$$, out of C and A is $$1$$−a and that of occurring three events simultaneously is $$a2$$, then the probability  that at least one out of A,B,C will occur is

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$$P($$ exactly one event out of A and B occurs $$)=PA$$∩B′∪A′∩$$B=P(A$$∪$$B)$$−$$P(A$$∩$$B)=P(A)+P(B)$$−$$2P(A$$∩$$B)$$∴ $$P(A)+P(B)$$−$$2P(A$$∩$$B)=1$$−$$a----1$$ Similarly, $$P(B)+P(C)$$−$$2P(B$$∩$$C)=1$$−$$2a----2P(C)+P(A)$$−$$2P(C$$∩$$A)=1$$−$$a----3P(A$$∩B∩$$C)=a2---4now$$ ,$$P(A$$∪B∪$$C)=P(A)+P(B)+P(C)$$−$$P(A$$∩$$B)$$−$$P(B$$∩$$C)$$−$$P(C$$∩$$A)+P(A$$∩B∩$$C)=12$$[$$P(A)+P(B)$$−$$2P(B$$∩$$C)+P(B)+P(C)$$−$$2P(B$$∩$$C)+P(C)+P(A)$$−$$2P(C$$∩$$A)$$]$$+P(A$$∩B∩$$C)=12$$[$$1$$−$$a+1$$−$$2a+1$$−a]$$+a2$$ [ using (1),(2),(3) and (4)] $$=32$$−$$2a+a2=12+(a$$−$$1)2$$>$$12$$

Let A,B,C be three events. If the probability of occurring exactly one event out of A and B is 1−a, out of B and C is 1−2a, out of C and A is 1−a and that of occurring three events simultaneously is a2, then the probability that at least one out of A,B,C will occur is

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