2. The profit from breaking the law in a transaction (e.g., selling a counterfei

Question

2. The profit from breaking the law in a transaction (e.g., selling a counterfeit iPad) is r. The odds of being caught is p. Find the expected profit of a merchant who performs n 2 3 illegal transactions in each of the following cases: (a) the merchant is issued a warning letter the first time they get caught, and is assessed a fine of 1 for each subsequent violation they get caught doing (b) the fine is 1 for the first three times the merchant gets caught, and then 10 for each subsequent violation that they get caught doing: (c) the fine is k for the -th violation that the merchant gets caught doing.

Explanation / Answer

Preparatory Work

Profit from n illegal transactions = r.n [given profit on breaking a law in a transaction is r] ...... (1)

Let out of n illegal transactions, the merchant is caught k times. Then, probability of this happening = nCkpk(1 - p)n – k ……………………………………………………….(2)

Thus, net expected profit from n transactions

= rn - [k =1,n]{(Fine for k times being caught) x P(being caught k times)}

= rn - [k =1,n]{(Fine for k times being caught) x nCkpk(1 - p)n – k}[vide (2) above]…… (3)

Back-up Theory

[k =1,n]{nCkpk(1 - p)n – k} = 1[being sum total of Binomial probabilities]…… ……….(4)

[k =1,n][k x {nCkpk(1 - p)n – k} = np [being mean Binomial Distribution]…… ……….(5)

Now, to work out the solution,

Part (a)

Expected profit = rn - [k =1,n]{(k – 1) x nCkpk(1 - p)n – k}[vide (3) above]

= rn - [k =1,n]{k x nCkpk(1 - p)n – k} + [k =1,n]{nCkpk(1 - p)n – k}

= rn – np + 1 [vide (4) and (5) above]

Thus, Expected profit = n(r – p) + 1 ANSWER   

Part (b)

Expected profit = rn - [k =1,n][{(3 x 1) + 10(k – 3)} x nCkpk(1 - p)n – k}] [vide (3) above]

= rn – 10np + 27 [vide (4) and (5) above]

Thus, Expected profit = n(r – 10p) + 27 ANSWER

Part (c)

Expected profit = rn - [k =1,n][k x nCkpk(1 - p)n – k}] [vide (3) above]

= rn – np [vide (4) and (5) above]

Thus, Expected profit = n(r – p) ANSWER

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