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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 ∪ = ∪ = ) ( , φ = ∩ 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 DeMorgan’s 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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 Spring '11
 voltz
 Probability, Probability theory, pn, Bernoulli trial

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