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Electricity power failure occurs on average 3 times every 20 weeks. What is the probability that there will not be more than one failure during a particular week?

$\lambda=3/20=$the average number of failures for any given week. $x=0,1$

$P(X=0)= ({e^{{-3}/{20}} \cdot ({3}/{20})^0})/{0!}=0.860707$

$P(X=1)= ({e^{{-3}/{20}} \cdot ({3}/{20})^1})/{1!}=0.129106$

$P(X\leq 1)=P(X=0)+P(X=1)=0.989814$

This answer makes sense since the average frequency of occurrence is very low. Is my solution correct?

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    Looks good! Your reasoning is correct. – 2017-02-22

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Yes, your answer and calculations are correct as shown, although I would be a bit clearer with notation and write $$\begin{align*} \Pr[X \le 1] &= \Pr[X = 0] + \Pr[X = 1] \\ &= e^{-3/20} \left(\frac{(3/20)^0}{0!} + \frac{(3/20)^1}{1!}\right) \\ &= e^{-3/20}\left(1 + \frac{3}{20}\right) \\ &= \frac{23}{20}e^{-3/20} \\ &\approx 0.989814. \end{align*}$$