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I am trying to le-learn probability theory and I am solving the following problem:

A probability that a manufactured device has $3\%$ or more deffects is $p=0.02$. If a company buys $5$ devices, what is the probability that at least one has $3\%$ or more defects.

I am thinking using binomial distribution to find probabilities that 1, 2, 3, 4 or all 5 devices are defective. So $$P(A)= {5\choose 1}p(1-p)^4+{5\choose 2}p^2(1-p)^3+{5\choose 3}p^3(1-p)^2+{5\choose 4}p^4(1-p)+{5\choose 5}p^5 $$

Is this a correct approach or there is an easier way to solve the problem?

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Yes, your approach is correct. An easier way to solve the problem is to use the converse probability.

$$P(X\geq 1)=1-P(X=0)=1-{5\choose 0}\cdot p^0\cdot (1-p)^5=1-(1-p)^5$$

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