If the rate of defaults per year in a set of companies is given by $\lambda = 5$ , what is the probability of four or more defaults in half a year?
A) 0.06
B) 0.12
C) 0.18
D) 0.24

Question

Quizplus · Accepted Answer

The number of defaults can be modeled using a Poisson distribution, where $\lambda$ is the average rate of occurrence. For half a year, the rate $\lambda = 5/2 = 2.5$ defaults. The probability of having $k$ defaults is given by $P(k) = \frac{e^{-\lambda} \lambda^k}{k!}$. To find the probability of four or more defaults, we calculate the probability of 0, 1, 2, and 3 defaults and subtract this sum from 1.$$P(0) = \frac{e^{-2.5} 2.5^0}{0!} = e^{-2.5}$$$$P(1) = \frac{e^{-2.5} 2.5^1}{1!} = 2.5e^{-2.5}$$$$P(2) = \frac{e^{-2.5} 2.5^2}{2!} = \frac{6.25e^{-2.5}}{2}$$$$P(3) = \frac{e^{-2.5} 2.5^3}{3!} = \frac{15.625e^{-2.5}}{6}$$Summing these probabilities gives the probability of having fewer than four defaults. Subtracting this sum from 1 gives the probability of having four or more defaults:$$1 - (P(0) + P(1) + P(2) + P(3)) = 1 - e^{-2.5}(1 + 2.5 + \frac{6.25}{2} + \frac{15.625}{6})$$Calculating this expression gives a value close to 0.24, making option D the correct answer.

If the Rate of Defaults Per Year in a Set λ=5\lambda = 5λ=5

If the Rate of Defaults Per Year in a Set $\lambda = 5$