Question

Intrusion detection systems. The Journal of Researchof the National Institute of Standards and Technology (November–December 2003) published a study of a doubleintrusion detection system with independent systems. Ifthere 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 withprobability .2, and system B sounds an alarm with probability.1. Now assume that the probability of an intruderis .4. Also assume that under a given condition (intruderor not), systems A and B operate independently. If bothsystems sound an alarm, what is the probability that anintruder is detected?

Short answer

The probability that anintruder is detected is 0.9661.

Step-by-step solution

Important formula

The formula for probability is$\mathbf{P}\mathbf{=}\frac{\mathbf{favourableoutcomes}}{\mathbf{totaloutcomes}}$.

The Baye’s formula is

$$\begin{gathered} \mathbf{P}\mathbf{\left(}\frac{{\mathbf{B}}_{\mathbf{i}}}{\mathbf{A}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\mathbf{⌒}\mathbf{A}\mathbf{)}}{\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{i}}\mathbf{\right)}\mathbf{P}\mathbf{\left(}{\displaystyle \frac{A}{B_i}}\mathbf{\right)}}{\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{1}}\mathbf{\right)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{i}}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{2}}\mathbf{\right)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{k}}\mathbf{\right)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{k}}}\mathbf{\right)}} \end{gathered}$$

The probability that anintruder is detected

Let,

A=intruder

B=system A sounds alarm

C=system B sounds alarm

$$\begin{gathered} P\left(A\right)=0.4 \\ P\left(B\backslash A\right)=0.9 \\ P\left(C\backslash A\right)=0.95 \\ P\left(B\backslash A^c\right)=0.2 \\ P\left(C\backslash A^c\right)=0.1 \end{gathered}$$

The systems are independent.

$$\begin{gathered} P\left(B⌒C\backslash A\right)=0.9\times 0.95 \\ =0.855 \end{gathered}$$

$$\begin{gathered} P\left(B⌒A\backslash A^c\right)=0.2\times 0.1 \\ =0.02 \end{gathered}$$

By using Baye’s rule, if both systems sound an alarm, the probability that an intruder is detected obtained as:

$$\begin{gathered} P\left(A\backslash B⌒C\right)=\frac{0.855\times 0.4}{0.855\times 0.4+0.02\times 0.6} \\ =0.9661 \end{gathered}$$

Therefore, the likelihood of spotting an intruder is 0.9661.