JEE Mains · Maths · STD 11 - 14. probability
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 \(\frac{7}{10^{7}}\)
- B \(\frac{81}{10^{8}}\)
- C \(\frac{67}{10^{8}}\)
- D \(\frac{73}{10^{8}}\)
Answer & Solution
Correct Answer
(D) \(\frac{73}{10^{8}}\)
Step-by-step Solution
Detailed explanation
10 defective & 90 non-defective Req. probability = (7 def 1 fair) or (8 defective) Req. probability \(=\frac{\left(10^7 \times 90\right) \times 8+10^8}{100^8}\) \(=\frac{72 \times 10^8+10^8}{100^8}=\frac{73}{10^8}\)
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