According to the Mortgage Bankers Association, 11% of U.S. mortgages were delinquent last year. A delinquent mortgage is one that has missed at least one payment but has not yet gone to foreclosure. A random sample of eight mortgages was selected. What is the probability that less than three of these mortgages are delinquent?

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This problem can be solved using the binomial probability formula, which is P(X = k) = (nCk) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful trials, p is the probability of success on a single trial, and nCk is the combination of n items taken k at a time. Here, n = 8 (the number of mortgages), p = 0.11 (the probability of a mortgage being delinquent), and we want the probability of k being less than 3, which means we need to calculate the probability for k = 0, 1, and 2, and then sum those probabilities.Calculating for k = 0, 1, and 2 and summing those probabilities gives a result close to 0.9794, indicating that there's a 97.94% chance that less than three of the selected mortgages are delinquent.