Probability of an egg being defective `= 10/100 = 1/10`
So, probability of an egg being non-defective` = 1-0.1=0.9`
`10` eggs are drawn successively with replacement.
So, the probability of getting no defective egg ` = (0.9)^10`
Hence, the probability that there is at least one defective egg = `1-(0.9)^10.`