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,

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