CUET · MATHS · PYQ PAPER 2023
Ten bulbs are drawn successively with replacement from a lot containing \(10 \%\) defective bulbs. The probability that there is at least one defective bulb is :
- A \(1-\left(9^9 / 10^9\right)\)
- B \(9 / 10\)
- C \(1-\left(9^{10} / 10^{10}\right)\)
- D \(1-\left(10^9 / 9^{10}\right)\)
Answer & Solution
Correct Answer
(C) \(1-\left(9^{10} / 10^{10}\right)\)
Step-by-step Solution
Detailed explanation
P(defective) \( = 0.1 \) P(not defective) \( = 1 - 0.1 = 0.9 \) P(no defective in 10 draws) \( = (0.9)^{10} = \left(\frac{9}{10}\right)^{10} = \frac{9^{10}}{10^{10}} \) P(at least one defective) \( = 1 - P(\text{no defective in 10 draws}) = 1 - \frac{9^{10}}{10^{10}} \)
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