What is the probability of six or more loans default?

Question

A bank has 100 one-year loans, each with a 1% probability of default. Assume that the probability of default applicable to all loans is uncertain. It is equally likely to be 0.5% or 1.5%. What is the probability of six or more defaults?

Answer 1

This is actually a probability question, and can be solved using a binomial distribution equation. For any given number of defaults (x) you can find out the probability of that number of defaults using this equation:

N! / ((N-x)! * x!) * p^x * (1-p)^(N-x)

N is the number of loans, p is the probability of default.

We could calculate the probability of all the possible default numbers that are 6 or greater, but it would be faster to calculate all the ones 5 or lower and then subtract that from 1.

When x = 0, 100! / ((100! * 0!) * 0.01⁰ * 0.99¹⁰⁰ = 0.366 or 36.6%.

Repeat these for x values between 1 & 5, add them all up, and we get 0.99946546554, meaning there is a 99.946546554% chance of having 5 or fewer defaults. If we subtract this probability from 100%, or 1, we will find the probability for 6 or more defaults: 0.00053453446, or 0.05%.