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 0.0130%. 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

Answer 1

0.000974

Explanation:

Let assume that;

P(Q) is the probability that the same user originated calls from two or more metropolitan areas in a single day

Also; Let consider M to be the event that denotes the legitimate users and N to be the event that denote the fraudulent users .

Then;

P(M) = 0.00013

P(N) = 1 - P(M)

P(N) = 1 - 0.00013

P(N) = 0.99987

P(Q|M) = 0.3

P(Q|N) = 0.4

The probability the same users originates calls from two or more metropolitan areas in a single day is calculated as follows:

P(Q) = (P(M) P(Q|M) ) + ( P(N) P(Q|N) )

P(Q) = ( 0.00013 × 0.3 ) + (0.99987 × 0.04 )

P(Q) = 0.000039 + 0.0399948

P(Q) = 0.0400338

However; 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,

`P(M|Q) = (P(M) P(Q|M))/((P(M) \ P(Q|M) ) + ((P(N) \ P(Q|N)) ) \\ \\ \\ P(M|Q) = (0.0001 * 0.3)/(0.0400338) \\ \\ \\ P(M|Q) = (0.000039)/(0.0400338) \\ \\ \\ \mathbfP(M`

Answer 2

0.0009741

Explanation:

The approach to solve this question is by the use of Baye’s theorem in conditional probability

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