Answer to Question #276567 in Statistics and Probability for Pere

Question

Software to detect fraud in consumer phone cards tracks the number of metropolitan areas where calls originate each day. It is found that 4% 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 0.0177%. 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?Show working please

Answer 1

The probability P(E) that the same users originates calls from two or more metropolitan areas in a single day s obtained below:

Let event A is denoted as users legitimate one and the event B is denoted as users fraudulent one.

Event E denotes selecting same users originates calling from the two or more metropolitans.

The proportion of fraudulent users is 0.01% and the probability of legitimate users is 99.99%.

That is, P(A)=0.0001 and P(B)=0.9999.

The probability that the fraudulent users original calls from two or more metropolitan city is 30% and the probability of the legitimate users originate from two or more metropolitan city is 1%. That is, $P(E \mid A)=0.3$ and $P(E \mid B)=0.01$

The probability the same users originates calls from two or more metropolitan areas in a single day is,

$P(E)=[P(A) P(E \mid A)]+[P(B) P(E \mid B)]\\ =0.00003+0.00999\\ =0.01002$

The probability that the users is fraudulent given that the same users originates calls from two or more metropolitan areas in a single day is,

$\begin{gathered} P(A \mid E)=\frac{P(A) P(E \mid A)}{P(A) P(E \mid A)+P(B) P(E \mid B)} \\ =\frac{0.0001(0.3)}{(0.0001 \times 0.3)+(0.9999 \times 0.01)} \\ =\frac{0.00003}{0.00003+0.009999} \\ =0.0029 \end{gathered}$