Question

Software to detect fraud in consumer phone cards tracks the number of metropolitan areas where calls originate each day. It is found that 2% 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 0.0121%. 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? Round your answer to six decimal places

Answer 1

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