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Example 32 Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is atleast one defective egg.Let X : be the number of defective eggs Picking eggs with replacement is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx π^(πβπ) π^π Here, n = number of eggs picked = 10 p = Probability of getting defective egg = 10% = 10/100 = 1/10 q = 1 β p = 1 β 1/10 = 9/10 Hence, P(X = x) = 10Cx (π/ππ)^π (π/ππ)^(ππ β π) We need to find Probability that there is atleast one defective egg P(atleast one defective egg) = 1 β P(getting 0 defective eggs) = 1 β P(X = 0) = 1 β 10C0(1/10)^0 (9/10)^(10 β0) = 1 β 1 Γ 1 Γ (9/10)^10 = 1 β (9/10)^10