Probability of failures

In certain productions, the probability of failures is 0.01. Calculate the probability that there will be more than 1 failure among the 100 selected products if we return the selected products to the file after the check.

Result

p =  0.264

Solution:

C0(100)=(1000)=100!0!(1000)!=11=1 C1(100)=(1001)=100!1!(1001)!=1001=100 q=0.01 n=100 p0=(n0) q0 (1q)n0=1 0.010 (10.01)10000.366 p1=(n1) q1 (1q)n1=100 0.011 (10.01)10010.3697  p=1(p0+p1)=1(0.366+0.3697)0.26420.264C_{{ 0}}(100) = \dbinom{ 100}{ 0} = \dfrac{ 100! }{ 0!(100-0)!} = \dfrac{ 1 } { 1 } = 1 \ \\ C_{{ 1}}(100) = \dbinom{ 100}{ 1} = \dfrac{ 100! }{ 1!(100-1)!} = \dfrac{ 100 } { 1 } = 100 \ \\ q=0.01 \ \\ n=100 \ \\ p_{0}={ { n } \choose 0 } \cdot \ q^0 \cdot \ (1-q)^{ n-0 }=1 \cdot \ 0.01^0 \cdot \ (1-0.01)^{ 100-0 } \doteq 0.366 \ \\ p_{1}={ { n } \choose 1 } \cdot \ q^1 \cdot \ (1-q)^{ n-1 }=100 \cdot \ 0.01^1 \cdot \ (1-0.01)^{ 100-1 } \doteq 0.3697 \ \\ \ \\ p=1-(p_{0}+p_{1})=1-(0.366+0.3697) \doteq 0.2642 \doteq 0.264



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