MATH SOLVE

2 months ago

Q:
# 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 0.0131%. 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).

Accepted Solution

A:

Answer:0.999987Step-by-step explanation:Given thatThe user is a legitimate one = E₁The user is a fraudulent one = E₂The same user originates calls from two metropolitan areas = AUse Bay's Theorem to solve the problemP(E₁) = 0.0131% = 0.000131P(E₂) = 1 - P(E₁) = 0.999869P(A/E₁) = 3% = 0.03P(A/E₂) = 30% = 0.3Given a randomly chosen user originates calls from two or more metropolitan, The probability that the user is fraudulent user is :[tex]P(E_2/A)=\frac{P(E_2)\times P(A/E_2)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}[/tex][tex]=\frac{(0.999869)(0.3)}{(0.000131)(0.03)+(0.999869)(0.3)}[/tex][tex]\frac{0.2999607}{0.00000393+0.2999607}[/tex][tex]\frac{0.2999607}{0.29996463}[/tex]= 0.999986898 ≈ 0.999987