[shahkb4]

shahkb4

3 June, 2026 at 8:27 pm #1124

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)