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software that can be used to detect fraund in customer
phonecards, tracks the number of metropolitan areas from which
callsoriginate each day. suppose 1% of the legitimate users
originatecalls from two or more metropolitan areas in a single day;
and 30%of fradulent users originate calls from two or more
metropolitanarea in a single day. The overall proportion of
fraudulent users is0.01%. If the same person originates calls from
two or moremetropolitan areas in as ingle, what is probability that
such useis fraudlent?

Respuesta :

Answer:

0.002991

Step-by-step explanation:

Let

User is legitimate one = E₁

User is fraudulent one = E₂

User originates calls from two metropolitan = A

Therefore, according to the question

P(E₁) = 1 - 0.01% = 1 - 0.0001 = 0.9999

P(E₂) = 0.0001

P(A/E₁) = 1% = 0.01

P(A/E₂) = 30% = 0.30

Now,

P(E₂/A) = [tex]\frac{P(E_2)\timesP(A/E_1)}{P(E_1)\timesP(A/E_1)+P(E_2)\timesP(A/E_2)}[/tex]

or

P(E₂/A) = [tex]\frac{0.9999\times0.01}{0.0001\times0.01+0.0001\times0.3}[/tex]

or

P(E₂/A) = 0.002991

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