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%=90100=0.9   C2(8)=(82)=8!2!(8−2)!=8⋅72⋅1=28  p2=(82)⋅ q2⋅ (1−q)8−2=28⋅ 0.92⋅ (1−0.9)8−2≐0  C3(8)=(83)=8!3!(8−3)!=8⋅7⋅63⋅2⋅1=56  p3=(83)⋅ q3⋅ (1−q)8−3=56⋅ 0.93⋅ (1−0.9)8−3≐0.0004  C4(8)=(84)=8!4!(8−4)!=8⋅7⋅6⋅54⋅3⋅2⋅1=70  p4=(84)⋅ q4⋅ (1−q)8−4=70⋅ 0.94⋅ (1−0.9)8−4≐0.0046  C5(8)=(85)=8!5!(8−5)!=8⋅7⋅63⋅2⋅1=56  p5=(85)⋅ q5⋅ (1−q)8−5=56⋅ 0.95⋅ (1−0.9)8−5≐0.0331  C6(8)=(86)=8!6!(8−6)!=8⋅72⋅1=28  p6=(86)⋅ q6⋅ (1−q)8−6=28⋅ 0.96⋅ (1−0.9)8−6≐0.1488  C7(8)=(87)=8!7!(8−7)!=81=8  p7=(87)⋅ q7⋅ (1−q)8−7=8⋅ 0.97⋅ (1−0.9)8−7≐0.3826  C8(8)=(88)=8!8!(8−8)!=11=1  p8=(88)⋅ q8⋅ (1−q)8−8=1⋅ 0.98⋅ (1−0.9)8−8≐0.4305 p=p2+p3+p4+p5+p6+p7+p8=0+0.0004+0.0046+0.0331+0.1488+0.3826+0.4305≐1≐0.9999993q=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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