I need help with this easy statistics and probability question?

[barnett]

Software to detect fraud in consumer phone cards tracks the number of metropolitan areas where calls originate each day. It is found that 3% 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 0.0162%. 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 (e.g. 98.765432).

[mate]

P[fraudulent | different areas] ?
P[different areas | fraudulent ] = 0.30
P[different areas | legitimate ] = 0.03
P[ fraudulent ] = 0.0162
P[fraudulent|different areas]P[different areas]
= P[different areas|fraudulent] P[fraudulent]
= 0.30 * 0.000162 = 0.0000486
=> P[fraudulent|different areas] = 0.0000486/P[different areas]
So now we focus on calculating P[different areas]
P[different areas] = P[different areas AND fraudulent]
+ P[different areas AND legitimate]
= 0.0000486 + 0.03*(1-0.000162)
= 0.03004374
=> P[fraudulent|different areas] = 0.0000486/0.03004374
= 0.00161764

This is an application of Bayes' rule. Switching a priori and a posteriori odds.

[fiynn]

F = Fraudulant calls
P(F)=.000162
P(F') = 1-.000162 =.999838

L = legigimate caller
C = fradulant caller
O =originates calls from two or more metropolitan areas in a single day
P(O/F) = .30
P(O/L)=.03

Apply Bayes' theorem:
P(F/O) = P(F)P(O/F) / [P(F)P(O/F)+ P(F')P(O/L)]
= (.000162)(.30) / [(.000162)(.30) + (.999838)(.03)]

=0.001618