From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

Q. From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

(A) $\dfrac{73}{10^8}$

(B) $\dfrac{67}{10^8}$

(C) $\dfrac{7}{10^7}$

(D) $\dfrac{81}{10^8}$

Correct Answer: $\dfrac{73}{10^8}$

Explanation

Probability that a bulb selected is defective is

$$ p=\frac{10}{100}=\frac{1}{10}, \quad q=\frac{9}{10} $$

Since selection is with replacement, the number of defective bulbs follows a binomial distribution with $n=8$.

Required probability is getting at least 7 defective bulbs:

$$ P(X\ge7)=P(X=7)+P(X=8) $$

$$ P(X=7)=\binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right) $$

$$ =\frac{72}{10^8} $$

$$ P(X=8)=\binom{8}{8}\left(\frac{1}{10}\right)^8=\frac{1}{10^8} $$

$$ P(X\ge7)=\frac{72}{10^8}+\frac{1}{10^8}=\frac{73}{10^8} $$

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