Answer:
the probability that the user is fraudulent is 0.00299133
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//](https://img.qammunity.org/2020/formulas/mathematics/college/h6nxzmj8pk74zjytrvhffa0x3ptjb13bbk.png)
With these values, the probability that a user is fraudulent, if it has originated calls from two or more areas is:
![P (F|G) = (P(F\bigcap G))/(P(G)) = (P(F)P(G|F))/(P(G)) = (P(F)P(G|F))/(P(F)P(G|F)+P(L)P(G|L))](https://img.qammunity.org/2020/formulas/mathematics/college/9lc63rtc6k4w2v244dfwayf0jozbzeuxvm.png)
![((0.0001)(0.30))/((0.0001)(0.30)+(0.9999)(0.01)) = 0.00299133](https://img.qammunity.org/2020/formulas/mathematics/college/f3i4jp907k1kqqo2xpz2qz1r0xo3o49le4.png)