Question

Suppose 6% of all people filing the long income tax form seek deductions that they know are illegal, and an additional 3% incorrectly list deductions because they are unfamiliar with income tax regulations. Of the 6% who are guilty of cheating, 76% will deny knowledge of the error if confronted by an investigator.
(a) If the filer of the long form is confronted with an unwarranted deduction and he or she denies the knowledge of the error, what is the probability that he or she is guilty?

Answer 1

Answer:

The probability that a person is guilty given that he or she  denies the knowledge of the error is 0.6068.

Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event E$_{i}$, belonging to the sample space S, given that another event A has already occurred is:

$P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{P(A|E_{1})P(E_{1})+P(A|E_{2}P(E_{2})+...+P(A|E_{n}P(E_{n})}$

Denote the events as follows:

X = illegal deduction is filed

Y = knowledge of the error is denied.

The information given is:

P (Cheating) = 0.06

P (Actual error) = 0.03

P (Y|X) = 0.76

Compute the probability of X as follows:

$P(X)=\frac{P(Cheating)}{P(Cheating)+P(Actual\ error)}=\frac{0.06}{0.06+0.03}=0.67$

The probability that a person who is not guilty will deny the knowledge of the error, is:

$P (Y|X^{c})=1$

Compute the value of P (X|Y) as follows:

$P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}$

             $=\frac{0.76\times 0.67}{(0.76\times 0.67)+(1\times (1-0.67))}$

             $=\frac{0.5092}{0.5092+0.33}$

             $=0.6068$

Thus, the probability that a person is guilty given that he or she  denies the knowledge of the error is 0.6068.