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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
[JEE MAIN 24 Jan 2026 S I]
(1) $\frac{7}{10^{7}}$
(2) $\frac{81}{10^{8}}$
(3) $\frac{67}{10^{8}}$
(4) $\frac{73}{10^{8}}$Solution
(10 Defective, 90 Non-defective)
Required Probability of getting atleast 7 defective bulbs out 8 bulbs $=\mathrm{P}($ Exactly 7 defective bulbs $)+\mathrm{P}($ all 8 defective bulbs $)$
$={ }^{8} \mathrm{C}_{7}\left(\frac{10}{100}\right)^{7}\left(\frac{90}{100}\right)+\left(\frac{10}{100}\right)^{8}=8\left(\frac{1}{10}\right)^{7}\left(\frac{9}{16}\right)+\left(\frac{1}{100}\right)^{8}=\frac{72+1}{108}=\frac{73}{10^{8}}$ Answer(4)
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