Question

If electricity power failures occur according to a Poisson distribution
with an average of 3 failures every twenty weeks, calculate the
probability that:
i. at most one failure during a particular week.

ii. exactly 4 failures within ten weeks.

Answer 1

Answer on Question #64235 – Math – Statistics and Probability

Question

If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that:

i. at most one failure during a particular week.

ii. exactly 4 failures within ten weeks.

Solution

i. Let the amount of the power failures during 1 week be $L_{1}$. The distribution of a random variable $L_{1}$ is a Poisson distribution with the average $\lambda_{1} = \lambda$:

$$P(L_{1} = l) = \frac{\lambda_{1}^{l}}{l!} e^{-\lambda_{1}}.$$

There are 3 failures during 20 weeks on average: $20E(L_{1}) = 3$,

hence $E(L_{1}) = \lambda_{1} = 3/20$ ($E(L_{1})$ is the mathematical expectation of $L_{1}$, i.e., its average).

Next,

$$\begin{array}{l} P(L_{1} \leq 1) = P(L_{1} = 0) + P(L_{1} = 1) = \frac{\lambda_{1}^{0}}{0!} e^{-\lambda_{1}} + \frac{\lambda_{1}^{1}}{1!} e^{-\lambda_{1}} = \frac{(\lambda t)^{0}}{0!} e^{-\lambda t} + \frac{(\lambda t)^{1}}{1!} e^{-\lambda t} = \\ = e^{-\lambda t} + \lambda t e^{-\lambda t} = (1 + \lambda t) e^{-\lambda_{1} t} = \left(1 + \frac{3}{20} \cdot 1\right) e^{-3/20} = \frac{23}{20} e^{-3/20} \approx 0.990. \end{array}$$

ii. Let the amount of the power failures during 10 weeks be $L_{10}$. The distribution of a random variable $L_{10}$ is a Poisson distribution with the average $\lambda_{10} = \lambda t$:

$$P(L_{10} = l) = \frac{\lambda_{10}^{l}}{l!} e^{-\lambda_{10}},$$

hence

$$P(L_{10} = 4) = \frac{(\lambda t)^{4}}{4!} e^{-\lambda t} = \frac{\left(\frac{3}{20} \cdot 10\right)^{4}}{4!} e^{-\frac{3}{20} \cdot 10} = \frac{\left(\frac{3}{2}\right)^{4}}{4!} e^{-\frac{3}{2}} = \frac{27}{128} e^{-\frac{3}{2}} \approx 0.047.$$

Indeed, the amount of the power failures during 20 weeks is a sum of 2 random variables with the mass function $\frac{\lambda_{10}^{l}}{l!} e^{-\lambda_{10}}$. Consequently, $2E(L_{10}) = 3$, $E(L_{10}) = \lambda_{10} = 3/2$.

Next,

$$P(L_{10} = 4) = \frac{\lambda_{10}^{4}}{4!} e^{-\lambda_{10}} = \frac{\left(\frac{3}{2}\right)^{4}}{4!} e^{-\frac{3}{2}} = \frac{27}{128} e^{-\frac{3}{2}} \approx 0.047$$

Answer:

i. The probability that at most one failure during a particular week is $\frac{23}{20} e^{-3/20} \approx 0.990$.

ii. The probability that exactly 4 failures within ten weeks is $\frac{27}{128} e^{-\frac{3}{2}} \approx 0.047$.

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