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Triss [41]
3 years ago
5

Five Cs Bank issues credit cards to its customers. In the past 3.4% of the out-of-state nonfraudulent credit card transactions w

ere made at gasoline stations and 9.2% of the out-of-state fraudulent credit card transactions were made at gasoline stations. Based on historical transactions 97.7% of out-of-state credit card transactions are legitimate while the rest of its out-of-state credit card transactions are fraudulent. An out-of-state credit card transaction took place at a gasoline station, what is the probability the transaction is fraudulent?
Mathematics
2 answers:
kolezko [41]3 years ago
7 0

Answer:

0.0838 (8.62%)

Step-by-step explanation:

defining the event G= an out-of-state transaction took place in a gasoline station , then the probability is

P(G) = probability that the transaction is fraudulent * probability that took place in a gasoline station given that is fraudulent + probability that the transaction is not fraudulent * probability that took place in a gasoline station given that is not fraudulent =  0.033 * 0.092 + 0.977 * 0.034 = 0.0362

then we use the theorem of Bayes for conditional probability. Defining also the event F= the transaction is fraudulent  , then

P(F/G)=P(F∩G)/P(G) =  0.033 * 0.092 /0.0362 =  0.0838 (8.62%)

where

P(F∩G)= probability that the transaction is fraudulent and took place in a gasoline station

P(F/G)= probability that the transaction is fraudulent given that it took place in a gasoline station

Slav-nsk [51]3 years ago
6 0

Answer:

Step-by-step explanation:

  • let F = fraud credit card transaction and G = legit credit card transaction
  • C =credit card transaction at gasoline station
  • P(G) = 0.977
  • P(F) = 1 - P(G) = 0.023
  • P(C|G) = 0.034
  • P(C|F) = 0.092

To find what is the probability the transaction is fraudulent = P(F|C)

From application of baye's rule ;

P(F|C) = P(F) P(C|F) / P(F) P(C|F) + P(G) P(C|G)

= 0.092 X 0.023 / 0.023 X 0.092 + 0.097 X 0.034

= 0.0599 is the probability the transaction is fraudulent

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