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Below is a depiction of the probability density functions of an IDS. The overlapping area of the two probability density functions represents the region in which there is the potential for false positives and false negatives. Suppose there is 1 actual intrusion for every 1000 authorized users, and the overlapping area covers 1% of the authorized users and 50% of the intruders. What is the probability that an event that occurs in this region is that of an authorized user

Respuesta :

Using conditional probability, it is found that there is a 0.00002 = 0.002% probability that an event that occurs in this region is that of an authorized user .

Conditional Probability

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Event in this region.
  • Event B: Authorized user.

Probability of an event in this region:

  • 1% of authorized users, which are 1 in 1000.
  • 50% of intruders, which are 999 in 1000.

Hence:

[tex]P(A) = 0.01\frac{1}{1000} + 0.5\frac{999}{1000} = \frac{0.01 + 0.5(999)}{1000} = \frac{499.51}{1000}[/tex]

The probability of an event and an authorized user is:

[tex]P(A \cap B) = 0.01\frac{1}{1000} = \frac{0.01}{1000}[/tex]

Hence, the conditional probability is:

[tex]P(B|A) = \frac{\frac{0.01}{1000}}{\frac{499.51}{1000}} = \frac{0.01}{499.51} = 0.00002[/tex]

0.00002 = 0.002% probability that an event that occurs in this region is that of an authorized user .

A similar problem is given at https://brainly.com/question/14398287

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