The probability of detecting a defective pistons in one test is "p=0.1" then using Bernoully formula "p=p_{10}(0)+p_{10}(1)+p_{10}(2)" where
"p_{10}(0)=\\frac{10!}{0!10!}0.1^00.9^{10}\\approx0.35 , p_{10}(1)=\\frac{10!}{1!9!}0.1^10.9^9\\approx0.39"
"p_{10}(2)=\\frac{10!}{2!8!}0.1^20.9^8\\approx0.19" then "p=0.93"
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