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 % = \frac{90}{100} = 0.9 C_{2} (8) = (\binom{8}{2}) = \frac{8!}{2! (8 - 2)!} = \frac{8 \cdot 7}{2 \cdot 1} = 28 p_{2} = (\binom{8}{2}) \cdot q^{2} \cdot (1 - q)^{8 - 2} = 28 \cdot 0. 9^{2} \cdot (1 - 0.9)^{8 - 2} ≐ 2.268 \cdot 1 0^{- 5} C_{3} (8) = (\binom{8}{3}) = \frac{8!}{3! (8 - 3)!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56 p_{3} = (\binom{8}{3}) \cdot q^{3} \cdot (1 - q)^{8 - 3} = 56 \cdot 0. 9^{3} \cdot (1 - 0.9)^{8 - 3} ≐ 0.0004 C_{4} (8) = (\binom{8}{4}) = \frac{8!}{4! (8 - 4)!} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70 p_{4} = (\binom{8}{4}) \cdot q^{4} \cdot (1 - q)^{8 - 4} = 70 \cdot 0. 9^{4} \cdot (1 - 0.9)^{8 - 4} ≐ 0.0046 C_{5} (8) = (\binom{8}{5}) = \frac{8!}{5! (8 - 5)!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56 p_{5} = (\binom{8}{5}) \cdot q^{5} \cdot (1 - q)^{8 - 5} = 56 \cdot 0. 9^{5} \cdot (1 - 0.9)^{8 - 5} ≐ 0.0331 C_{6} (8) = (\binom{8}{6}) = \frac{8!}{6! (8 - 6)!} = \frac{8 \cdot 7}{2 \cdot 1} = 28 p_{6} = (\binom{8}{6}) \cdot q^{6} \cdot (1 - q)^{8 - 6} = 28 \cdot 0. 9^{6} \cdot (1 - 0.9)^{8 - 6} ≐ 0.1488 C_{7} (8) = (\binom{8}{7}) = \frac{8!}{7! (8 - 7)!} = \frac{8}{1} = 8 p_{7} = (\binom{8}{7}) \cdot q^{7} \cdot (1 - q)^{8 - 7} = 8 \cdot 0. 9^{7} \cdot (1 - 0.9)^{8 - 7} ≐ 0.3826 C_{8} (8) = (\binom{8}{8}) = \frac{8!}{8! (8 - 8)!} = \frac{1}{1} = 1 p_{8} = (\binom{8}{8}) \cdot q^{8} \cdot (1 - q)^{8 - 8} = 1 \cdot 0. 9^{8} \cdot (1 - 0.9)^{8 - 8} ≐ 0.4305 p = p_{2} + p_{3} + p_{4} + p_{5} + p_{6} + p_{7} + p_{8} = 2.268 \cdot 1 0^{- 5} + 0.0004 + 0.0046 + 0.0331 + 0.1488 + 0.3826 + 0.4305 = 0.9999993

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