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There has been a great deal of controversy over the last several years regarding what types of surveillance are appropriate to prevent terrorism. Suppose a particular surveillance system has a 96% chance of correctly identifying a future terrorist and a 99.8% chance of correctly identifying someone who is not a future terrorist. If there are 1,000 future terrorists in a population of 200 million, and one of these 200 million is randomly selected, scrutinized by the system, and identified as a future terrorist, what is the probability that he/she actually is a future terrorist

Sagot :

Answer:

0.9976 = 99.76% probability that he/she actually is a future terrorist

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[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 question:

Event A: Person identified as a future terrorist

Event B: Person is actually a future terrorist.

Probability that a person is identified as a future terrorist:

96% of [tex]\frac{1000}{200000000}[/tex]

100 - 99.8 = 0.2% of [tex]\frac{200000000-1000}{200000000}[/tex]

So

[tex]P(A) = 0.96\frac{1000}{200000000} + 0.002\frac{200000000-1000}{200000000} = 0.002[/tex]

Probability of being identified and being a future terrorist:

100 - 99.8 = 0.2% of [tex]\frac{200000000-1000}{200000000}[/tex]

[tex]P(A \cap B) = 0.002\frac{200000000-1000}{200000000} = 0.00199999[/tex]

What is the probability that he/she actually is a future terrorist

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.00199999}{0.00200479} = 0.9976[/tex]

0.9976 = 99.76% probability that he/she actually is a future terrorist

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