Question

A pawn shop is open to purchasing goods that may not have been acquired through honest means. Every time the shop makes a questionable transaction there is a probability that the sale gets busted by the local police. The owner estimates that the probability that any single transaction will be busted is 2% independent from all others. He also estimates that the net profit he makes on one transaction is well-described by a normal distribution with average $50 with standard deviation of $15.

i. How much profit can they expect to see before the operation is busted?

ii. Find the standard deviation for total profit.

iii. A local police chief, for a monthly fee of $100, can make sure that they are not bothered as often, effectively reducing probability of a bust on each transaction to 0.5%. Suppose that they do 1 questionable transaction a day. Estimate whether they can expect to make more or less profit if they pay the bribe?

Answer 1

Probability of getting busted =P=0.02

a)

Let X is number of transaction before getting busted

So X is Geometric distribution with P=0.02

So

E(X)=(1-P)/P=0.98/0.02=49

Since we can expect 49 transaction before getting busted then expected profit =49*profit per transaction=49*50=2450

b)

Since we have standard deviation per transaction is 15 dollers

As calculated in part (a) we can expect 49 transaction before getting busted so

Standard deviation for total profit=square roo(49)*15

=7*15=105

C)

As there is 30 days in one month

So without paying bribe we expect 30*0.02=0.6 days police busted so shop can't get profit for 0.6 days so expected profit =29.4*50=1470

While if shop paid bribe to police then we expect 30*0.005=0.15days police busted so shop can't get profit for 0.15 days hence expected profit =50*29.85=1492.5

But there is bribe of 100 dollers then profit remain is 1392.5

So profit without giving bribe is more than profit with giving bribe.