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Unformatted text preview: 1 Lecture2 More on Independence and Bernoulli Trials Independence : Events A and B are independent if It is easy to show that A , B independent implies are all independent pairs. For example, and so that or i.e., and B are independent events. ). ( ) ( ) ( B P A P AB P = (1) ; , B A B A B A , ; , B A AB B A A B = = ) ( , = B A AB ), ( ) ( ) ( )) ( 1 ( ) ( ) ( ) ( ) ( B P A P B P A P B P A P B P B A P = = = ) ( ) ( ) ( ) ( ) ( ) ( ) ( B A P B P A P B A P AB P B A AB P B P + = + = = A 2 If P ( A ) = 0, then since the event always, we have and (1) is always satisfied. Thus the event of zero probability is independent of every other event! Independent events obviously cannot be mutually exclusive, since and A , B independent implies Thus if A and B are independent, the event AB cannot be the null set. More generally, a family of events are said to be independent, if for every finite subcollection we have A AB , ) ( ) ( ) ( = = AB P A P AB P ) ( , ) ( B P A P . ) ( AB P , , , , 2 1 n i i i A A A = = = n k i n k i k k A P A P 1 1 ). ( (2) { } i A 3 Let a union of n independent events. Then by DeMorgans law and using their independence Thus for any A as in (3) a useful result. , 3 2 1 n A A A A A = (3) n A A A A 2 1 = (4) . )) ( 1 ( ) ( ) ( ) ( 1 1 2 1 = = = = = n i i n i i n A P A P A A A P A P (5) , )) ( 1 ( 1 ) ( 1 ) ( 1 = = = n i i A P A P A P (6) 4 Example 2.1: Three switches connected in parallel operate independently. Each switch remains closed with probability p . (a) Find the probability of receiving an input signal at the output. (b) Find the probability that switch S 1 is open given that an input signal is received at the output. Solution: a. Let A i = Switch S i is closed. Then Since switches operate independently, we have 1 S 2 S 3 S Input Output , ) ( p A P i = ). ( ) ( ) ( ) ( ); ( ) ( ) ( 3 2 1 3 2 1 A P A P A P A A A P A P A P A A P j i j i = = . 3 1 = i 5 Let R = input signal is received at the output. For the event R to occur either switch 1 or switch 2 or switch 3 must remain closed, i.e., (7) (8) (9) . 3 2 1 A A A R = . 3 3 ) 1 ( 1 ) ( ) ( 3 2 3 3 2 1 p p p p A A A P R P + = = = ). ( )  ( ) ( )  ( ) ( 1 1 1 1 A P A R P A P A R P R P + = , 1 )  ( 1 = A R P 2 3 2 1 2 ) ( )  ( p p A A P A R P = = Using (3)  (6), We can also derive (8) in a different manner. Since any event and its compliment form a trivial partition, we can always write But and and using these in (9) we obtain , 3 3 ) 1 )( 2 ( ) ( 3 2 2 p p p p p p p R P + = + = (10) which agrees with (8). 6 Note that the events A 1 , A 2 , A 3 do not form a partition, since they are not mutually exclusive. Obviously any two or all three switches can be closed (or open) simultaneously....
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This note was uploaded on 04/24/2011 for the course ECE 6303 taught by Professor Voltz during the Spring '11 term at NYU Poly.
 Spring '11
 voltz

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