Answer:
The probability is 0.001812
Step-by-step explanation:
Let's call: F the event in which a user is fraudulent, L the event in which a user is legitimate and A the event in which the user calls from two or more metropolitan areas.
So, the probability that the user is fraudulent given that the user call from two or more metropolitan areas is calculated as:
P(F/A) = P(F∩A)/P(A)
Where P(A) = P(F∩A) + P(L∩A)
So, P(F∩A) is the probability that the user is fraudulent and the user calls from two or more metropolitan areas. it is calculated as:
P(F∩A) = 0.0121% * 30% = 0.00363%
Because 0.0121% is the proportion of fraudulent users and 30% is the percentage of the fraudulent users that call from two or more metropolitan areas in a single day.
At the same way, P(L∩A) is the probability that a user is legitimate and the users calls from two or more metropolitan areas. It is calculated as:
P(L∩A) = (100%-0.0121%) * 2% = 1.999758%
Because (100%-0.0121%) is the proportion of legitimate users and 2% is the percentage of the legitimate users that call from two or more metropolitan areas in a single day.
Then, P(A) is calculated as:
P(A) = 0.00363% + 1.999758%
P(A) = 2.003388%
Finally, If the same user originates calls from two or more metropolitan areas in a single day, the probability that the user is fraudulent, P(F/A), is equal to:
P(F/A) = 0.00363 / 2.003388
P(F/A) = 0.001812
