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10 works hard drawn successively with replacement from a lot containing 10% defective bulb. The probability that there is at least one defective bulb is :

Solution

✅ Correct Option: 3

Let XX = number of defective bulbs in 10 draws. p=0.1,q=0.9,n=10p = 0.1, q = 0.9, n = 10. P(at least one defective) =1−P(X=0)=1−(100)(0.9)10=1−(910)10= 1 - P(X = 0) = 1 - \binom{10}{0} (0.9)^{10} = 1 - \left(\frac{9}{10}\right)^{10}.

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