Deficiencies

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

Correct answer:

p =  95.003 %

Step-by-step explanation:

C0(10)=(010)=0!(100)!10!=11=1 C1(10)=(110)=1!(101)!10!=110=10 C2(10)=(210)=2!(102)!10!=21109=45 C3(10)=(310)=3!(103)!10!=3211098=120 q=2000300=203=0.15 n=10  p0=(0n) q0 (1q)n0=1 0.150 (10.15)1000.1969 p1=(1n) q1 (1q)n1=10 0.151 (10.15)1010.3474 p2=(2n) q2 (1q)n2=45 0.152 (10.15)1020.2759 p3=(3n) 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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