# Records

Records indicate 90% error-free. If 8 records are randomly selected, what is the probability that at least 2 records have no errors?

Result

p =  0.9999993

#### Solution:

$q=90 \%=\dfrac{ 90 }{ 100 }=0.9 \ \\ \ \\ \ \\ C_{{ 2}}(8)=\dbinom{ 8}{ 2}=\dfrac{ 8! }{ 2!(8-2)!}=\dfrac{ 8 \cdot 7 } { 2 \cdot 1 }=28 \ \\ \ \\ p_{2}={ { 8 } \choose 2 } \cdot \ q^2 \cdot \ (1-q)^{ 8-2 }=28 \cdot \ 0.9^2 \cdot \ (1-0.9)^{ 8-2 } \doteq 0 \ \\ \ \\ C_{{ 3}}(8)=\dbinom{ 8}{ 3}=\dfrac{ 8! }{ 3!(8-3)!}=\dfrac{ 8 \cdot 7 \cdot 6 } { 3 \cdot 2 \cdot 1 }=56 \ \\ \ \\ p_{3}={ { 8 } \choose 3 } \cdot \ q^3 \cdot \ (1-q)^{ 8-3 }=56 \cdot \ 0.9^3 \cdot \ (1-0.9)^{ 8-3 } \doteq 0.0004 \ \\ \ \\ C_{{ 4}}(8)=\dbinom{ 8}{ 4}=\dfrac{ 8! }{ 4!(8-4)!}=\dfrac{ 8 \cdot 7 \cdot 6 \cdot 5 } { 4 \cdot 3 \cdot 2 \cdot 1 }=70 \ \\ \ \\ p_{4}={ { 8 } \choose 4 } \cdot \ q^4 \cdot \ (1-q)^{ 8-4 }=70 \cdot \ 0.9^4 \cdot \ (1-0.9)^{ 8-4 } \doteq 0.0046 \ \\ \ \\ C_{{ 5}}(8)=\dbinom{ 8}{ 5}=\dfrac{ 8! }{ 5!(8-5)!}=\dfrac{ 8 \cdot 7 \cdot 6 } { 3 \cdot 2 \cdot 1 }=56 \ \\ \ \\ p_{5}={ { 8 } \choose 5 } \cdot \ q^5 \cdot \ (1-q)^{ 8-5 }=56 \cdot \ 0.9^5 \cdot \ (1-0.9)^{ 8-5 } \doteq 0.0331 \ \\ \ \\ C_{{ 6}}(8)=\dbinom{ 8}{ 6}=\dfrac{ 8! }{ 6!(8-6)!}=\dfrac{ 8 \cdot 7 } { 2 \cdot 1 }=28 \ \\ \ \\ p_{6}={ { 8 } \choose 6 } \cdot \ q^6 \cdot \ (1-q)^{ 8-6 }=28 \cdot \ 0.9^6 \cdot \ (1-0.9)^{ 8-6 } \doteq 0.1488 \ \\ \ \\ C_{{ 7}}(8)=\dbinom{ 8}{ 7}=\dfrac{ 8! }{ 7!(8-7)!}=\dfrac{ 8 } { 1 }=8 \ \\ \ \\ p_{7}={ { 8 } \choose 7 } \cdot \ q^7 \cdot \ (1-q)^{ 8-7 }=8 \cdot \ 0.9^7 \cdot \ (1-0.9)^{ 8-7 } \doteq 0.3826 \ \\ \ \\ C_{{ 8}}(8)=\dbinom{ 8}{ 8}=\dfrac{ 8! }{ 8!(8-8)!}=\dfrac{ 1 } { 1 }=1 \ \\ \ \\ p_{8}={ { 8 } \choose 8 } \cdot \ q^8 \cdot \ (1-q)^{ 8-8 }=1 \cdot \ 0.9^8 \cdot \ (1-0.9)^{ 8-8 } \doteq 0.4305 \ \\ p=p_{2}+p_{3}+p_{4}+p_{5}+p_{6}+p_{7}+p_{8}=0+0.0004+0.0046+0.0331+0.1488+0.3826+0.4305 \doteq 1 \doteq 0.9999993$

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