Question

Fraud detection has become an indispensable tool for banks and credit card companies to combat fraudulent credit card transactions. A fraud detection firm has detected some form of fraudulent activities in 2%, and serious fraudulent activities in 0.75% of transactions. Assume that fraudulent transactions remain stable.
a. What is the probability that fewer than 2 out of 110 transactions are fraudulent?
b. What is the probability that fewer than 2 out of 105 transactions are seriously fraudulent?

Answer 1

Answer:

a) 35.17% probability that fewer than 2 out of 110 transactions are fraudulent

b) 81.35% probability that fewer than 2 out of 105 transactions are seriously fraudulent

Step-by-step explanation:

For each transaction, there are only two possible outcomes. Either they are fradulent(or seriously fraudulent), or they are not. Transactions are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

a. What is the probability that fewer than 2 out of 110 transactions are fraudulent?

2% are fraudulent, so $p = 0.02$

110 transactions, so $n = 110$

This is

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{110,0}.(0.02)^{0}.(0.98)^{110} = 0.1084$

$P(X = 1) = C_{110,1}.(0.02)^{1}.(0.98)^{109} = 0.2433$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1084 + 0.2433 = 0.3517$

35.17% probability that fewer than 2 out of 110 transactions are fraudulent.

b. What is the probability that fewer than 2 out of 105 transactions are seriously fraudulent?

0.75% are seriously fraudulent, so $p = 0.0075$

105 transactions, so $n = 105$

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{105,0}.(0.0075)^{0}.(0.9925)^{105} = 0.4536$

$P(X = 1) = C_{105,1}.(0.0075)^{1}.(0.9925)^{104} = 0.3599$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.4536 + 0.3599 = 0.8135$

81.35% probability that fewer than 2 out of 105 transactions are seriously fraudulent