SOLUTION: There is a .02 probability that a customer's Visa charge will be rejected at a certain Target store because the transaction exceeds the customer's credit limit. What is the expecte

Question 1197331: There is a .02 probability that a customer's Visa charge will be rejected at a certain Target store because the transaction exceeds the customer's credit limit. What is the expected number of Visa transactions until the first one is rejected?
Answer by ikleyn(52803) (Show Source): You can put this solution on YOUR website!
.
There is a .02 probability that a customer's Visa charge will be rejected
at a certain Target store because the transaction exceeds the customer's credit limit.
What is the expected number of Visa transactions until the first one is rejected?
~~~~~~~~~~~~~~~

P(rejected right at the 1st use)               = 0.02.

P(rejected at the 2nd use) =    = 

P(rejected at the 3rd use) =  = 

P(rejected at the 4th use) =  = 

P(rejected at the 5th use) =  = 

  . . . . . .                    . . . . . .            . . . . . . 


P(rejected at the nth use) =  = 

  . . . . . .                    . . . . . .            . . . . . . 


So, the Math expectation ME is this sum

    ME =  = 

       = 


It is well known fact that the sum     is   .


So, in our case  ME =  =  =  =  = 49.


ANSWER.  The expected number of Visa transactions until the first one is rejected is 49.

Solved.

Notice that the answer is consistent with the fact that 49 = - = 50 - 1, which confirms the answer.

-----------------

Regarding the formula for the sum = , below I deduce it.

Let 

    S = x + 2x^2 + 3x^3 + 4x^4 + . . .      (1)

Multiply by x both sides,

   xS =      x^2 + 2x^3 + 3x^4 + . . .      (2)


From equation (1), subtract equation (2).  You will get

   S(1−x) = x + x^2 + x^3 + x^4 + . . . 

   S(1−x) = 

   S      = .

QED.

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