Question

Software to detect fraud in consumer phone cards tracks the number of metropolitan areas where calls originate each day. It is found that 1% of the legitimate users originate calls from two or more metropolitan areas in a single day. However, 30% of fraudulent users originate calls from two or more metropolitan areas in a single day. The proportion of fraudulent users is 0.01%. If the same user originates calls from two or more metropolitan areas in a single day, what is the probability that the user is fraudulent?

Answer 1

Answer:

the probability that the user is fraudulent is 0.00299133

Step-by-step explanation:

Let be the events be:

G: The user generates calls from two or more areas.

NG: The user does NOT generate calls from two or more areas.

L: The user is legitimate.

F: The user is fraudulent.

The probabilities established in the statement are:

$P (G | L) = 0.01//P (G | F) = 0.30//P (F) = 0.0001//P (L) = 0.9999//$

With these values, the probability that a user is fraudulent, if it has originated calls from two or more areas is:

$P (F|G) = \frac{P(F\bigcap G)}{P(G)} = \frac{P(F)P(G|F)}{P(G)} = \frac{P(F)P(G|F)}{P(F)P(G|F)+P(L)P(G|L)}$

$\frac{(0.0001)(0.30)}{(0.0001)(0.30)+(0.9999)(0.01)} = 0.00299133$