The probability of detecting a defective pistons in one test is p=0.1p=0.1p=0.1 then using Bernoully formula p=p10(0)+p10(1)+p10(2)p=p_{10}(0)+p_{10}(1)+p_{10}(2)p=p10(0)+p10(1)+p10(2) where
p10(0)=10!0!10!0.100.910≈0.35,p10(1)=10!1!9!0.110.99≈0.39p_{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\approx0.39p10(0)=0!10!10!0.100.910≈0.35,p10(1)=1!9!10!0.110.99≈0.39
p10(2)=10!2!8!0.120.98≈0.19p_{10}(2)=\frac{10!}{2!8!}0.1^20.9^8\approx0.19p10(2)=2!8!10!0.120.98≈0.19 then p=0.93p=0.93p=0.93
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