Records

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

Correct result:

p =  0.9999993

Solution:

q=90%=90100=0.9   C2(8)=(82)=8!2!(82)!=8721=28  p2=(82) q2 (1q)82=28 0.92 (10.9)822.268105  C3(8)=(83)=8!3!(83)!=876321=56  p3=(83) q3 (1q)83=56 0.93 (10.9)830.0004  C4(8)=(84)=8!4!(84)!=87654321=70  p4=(84) q4 (1q)84=70 0.94 (10.9)840.0046  C5(8)=(85)=8!5!(85)!=876321=56  p5=(85) q5 (1q)85=56 0.95 (10.9)850.0331  C6(8)=(86)=8!6!(86)!=8721=28  p6=(86) q6 (1q)86=28 0.96 (10.9)860.1488  C7(8)=(87)=8!7!(87)!=81=8  p7=(87) q7 (1q)87=8 0.97 (10.9)870.3826  C8(8)=(88)=8!8!(88)!=11=1  p8=(88) q8 (1q)88=1 0.98 (10.9)880.4305 p=p2+p3+p4+p5+p6+p7+p8=2.268105+0.0004+0.0046+0.0331+0.1488+0.3826+0.4305=0.9999993



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