During the hygienic inspection in 2000 mass caterers, deficiencies were found in 300 establishments. What is the probability that deficiencies in a maximum of 3 devices will be found during the inspection of 10 devices?

Correct result:

p =  95.003 %


C0(10)=(100)=10!0!(100)!=11=1 C1(10)=(101)=10!1!(101)!=101=10 C2(10)=(102)=10!2!(102)!=10921=45 C3(10)=(103)=10!3!(103)!=1098321=120 q=3002000=320=0.15 n=10  p0=(n0) q0 (1q)n0=1 0.150 (10.15)1000.1969 p1=(n1) q1 (1q)n1=10 0.151 (10.15)1010.3474 p2=(n2) q2 (1q)n2=45 0.152 (10.15)1020.2759 p3=(n3) q3 (1q)n3=120 0.153 (10.15)1030.1298  p=100 (p0+p1+p2+p3)=100 (0.1969+0.3474+0.2759+0.1298)=95.003%

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