statistics and probablity

Question

According to the Association, 6% of mexican mortgages were delinquent last year. A delinquent mortgage is one that has missed at least one payment but hasnot yet gone to foreclosure. A random sample of 30 mortgages was selected. What is the probability that less than two of these mortgages are delinquent

Answer 1

Well, I am going to make some assumptions which are typical in probability questions, but not explicitly stated here.

I am going to assume that delinquency is independent, and that this random sample was taken from the population as a whole (that much seems to be clear).

Under these assumptions you are effectively dealing with a binomial distribution. If that is not ringing familiar, a binomial distribution models a situation where with some probability $p$ you encounter a success, independent of all other trials, and with probability $q = 1-p$ you encounter a "failure". In this case we can define $p = 0.06$. The binomial distribution states that for $k$ sucesses in $n$ trials you take ${n \choose k}\cdot p^k\cdot (1-p)^{n-k}$. This should make intuitive sense since, there are $k$ possible trials which could have been successes out of the $n$ of them (i.e. n choose k). The probability of $k$ successes is $p^k$ and the remaining probability is of course $(1-p)^{n-k}$. Think of binomial distributions as flipping a coin $n$ times, looking for $k$ heads, given a probability of heads $p = ...$

The probability that less than two of these is delinquent is the sum of two probabilities, namely that $0$ are delinquent and that $1$ is delinquent.

That gives:

\begin{align} P(E) &= P(0) + P(1) \\ &= (1-0.6)^{30} + {30 \choose 1} \cdot 0.06^{1} \cdot (1 - 0.06)^{29} \\ &= .2992... \end{align}