Pr(R|A)=0.99 Pr(R|A c )=0.1 The probability of false alarm (a false indication of aircraft presence) =Pr(R∩A c )=Pr(R| A c )Pr(A c )=0.1×0.95=0.095 . The probability of missed detection (nothing registers, even though an aircraft is present) =Pr(R c ∩A)=Pr(R c |A)Pr(A)=0.01×0.05=0.0005 . The probability that aircraft is present given that radar registers its presence =Pr(A|R)=Pr(R| A)Pr(A)Pr(R|A)Pr(A)+Pr(R|A c )Pr(A c )=0.99×0.050.99×0.05+0.1×0.95≈0.3426 . 10. On an average the number of hits on a certain web page are 20 hits per day. What is the probability of observing more than 30 hits in a day? Hint : Use Poisson Distribution Ans: Let ? denote the number of hits on a certain webpage in a day with average number of hits equal to 20 i.e. ?( ? )=20 . We can use Poisson distribution to model the uncertainty in such cases i.e. ? ∼Pois(20) . So the probability that ? is more than 30 is Pr( ? >30)= ∑ i x = 31 ∞ 20 x e − 20 x ! = 0.01347468 6
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- Summer '18
- Sagar Arora