Probability Problem: Defaulting on Payments

[war485]

Problem:
The probability of defaulting on the n^th payment is 0.017n - 0.013 where n is a whole number, n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
find:
1) the probability of making all 10 payments
2) the probability of making only the first 5 payments
Note: each payment made are equal

My idea:

I was thinking that defaulting means I won't make the payment, so
is this the right answer to the first question?
P = ( 1 - (0.017 - 0.013)) * ( 1 - (0.017(2) - 0.013)) * ( 1 - (0.017(3) - 0.013)) * ... * ( 1 - (0.017(10) - 0.013))

and similarly for the second one:
P = P = ( 1 - (0.017 - 0.013)) * ( 1 - (0.017(2) - 0.013)) * ( 1 - (0.017(3) - 0.013)) * ( 1 - (0.017(4) - 0.013)) * ( 1 - (0.017(5) - 0.013))

Does that make sense? Taking the probability for each payment by 1 - failure and then multiply each of them. Am I doing this right? I'm not good at probability yet.

 

[robert Ihnot]

You have the right idea. They are independent events. My TI-86 gets 42.6% possibility of not defaulting on ten payments.

A simple problem is to consider the possibility of test failure to be 1/4 and they are to be three tests. Then the possibiility of passing all tests is (3/4)^3.

 

[bpet]

It often helps to draw the probability tree - try this and you might spot the term missing from one of your expressions.

 

[war485]

It often helps to draw the probability tree - try this and you might spot the term missing from one of your expressions.

? I checked it over with a tree diagram. How am I missing a term?

(edit): would I have to include the probability of defaulting in the second question? i.e. keep multiplying my answer by (0.017n - 0.013) where n = 6, ... 10 ?

(edit #2): I got it now. Thanks guys!