- #1

war485

- 92

- 0

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.