Let X = number of defective bulbs in 10 draws. p=0.1,q=0.9,n=10. P(at least one defective) =1−P(X=0)=1−(010)(0.9)10=1−(109)10.
10 works hard drawn successively with replacement from a lot containing 10% defective bulb. The probability that there is at least one defective bulb…
CUET UG 2022 — Mathematics Applied-Mathematics
2022mcqmedium
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 :
Official previous-year question
Held on 11 Aug 2022 · Verified 13 Jul 2026.
Options
- A
1−(109)9
- B
109
- C
1−(109)10
- D
1−1010109
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