Question

3.54 Intrusion detection systems. Refer to the Journal of Research of the National Institute of Standards and Technology (Nov.-Dec. 2003) study of a double intrusion detection system with independent systems, Exercise 3.39 (p. 106). Recall that if there is an intruder, system A sounds an alarm with probability 9 and system B sounds an alarm with probability .95. If there is no intruder, system A sounds an alarm with probability .2 and system B sounds an alarm with probability .1. Now, assume that the probability of an intruder is .4. If both systems sound an alarm, what is the probability that an intruder is detected?

3.54!

Answer 1

It Intruder A = Systern A sounds an Alarm Syste tem B Sounds an alarm B P (A\I)=0.9 PCB I) = 0.95 PCI)=0.4 PCA \I')=0,2 PCB\])=0.) PCI'S=0.6 In order to find the probality that an intercoder is deteated, given that both systems sound an alarm, must find the probability of an intruder given the intersection of A & B. Because systems A & B operate w we independently, probabilities PO AXB) - PCIA) PCT 13) Since we know the conditional of each system soun ounding an alarm, given the mutually exclusive § exhaustive events I and I we can (це Baye's Rule

to find the probability of an intruder given probability of and intruder given Bi А us the PCI) PCAVI) PCJA) PCI) PCAVI)+PCT) P(411) 19 (0.4) (0.9) (0.4)(0.9)+(0.6) (0.2) 0.36 0.48 PCJA) = 0.75 P (I) PCBI) PCIXB) - PCI) P(BLI) + P (ISP (B\JC) (0.4) (0,95) (0.4)(0.95) + 0.6) (0.1) 0.38 0.44 P(INB) >0,86

thing into the By substitu these Value Original equation, PCI\ AB) = PCINA) PCIB) 6.75) C0.86) P (IAB) = 0.6457 :.. the probabi sility that an intruder is that both systems sound detected given alarm is 0.645. an