Ten eggs drawn with replacement from a lot of 10% defective. Probability that **at least one** is defective:
- $1 - (9/10)^{10}$
- $(1/10)^{10}$
- $10 \cdot (1/10)$
- $10!$
Show answer and solution
Correct answer: A. $1 - (9/10)^{10}$
$P(X \geq 1) = 1 - P(X = 0) = 1 - {}^{10}C_0 (1/10)^0 (9/10)^{10} = 1 - (9/10)^{10} \approx 1 - 0.349 = 0.651$.