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(D) Let $n$ be the number of parts produced. The probability of a part being defective is $p = 0.05 = \frac{1}{20}$.The probability of a part being no

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$14$

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(D) Let $n$ be the number of parts produced. The probability of a part being defective is $p = 0.05 = \frac{1}{20}$.The probability of a part being non-defective is $q = 1 - 0.05 = 0.95 = \frac{19}{20}$.We want the probability of at least one defective part to be at least $1/2$, i.e.,$P(X \geq 1) \geq 1/2$.This is equivalent to $1 - P(X = 0) \geq 1/2$, where $P(X = 0)$ is the probability that no parts are defective.$1 - (0.95)^n \geq 0.5 \implies 0.5 \geq (0.95)^n$.Taking $\log_{10}$ on both sides: $\log_{10}(0.5) \geq n \log_{10}(0.95)$.$-\log_{10}(2) \geq n(\log_{10}(95) - \log_{10}(100))$.$-0.3 \geq n(1.977 - 2)$.$-0.3 \geq n(-0.023)$.Since we are dividing by a negative number, the inequality sign reverses: $n \geq \frac{0.3}{0.023} = \frac{300}{23} \approx 13.04$.Since $n$ must be an integer, the smallest integer $n$ is $14$.

Suppose a machine produces metal parts that contain some defective parts with probability $0.05$. How many parts should be produced in order that the probability of at least one part being defective is at least $1/2$? (Given that, $\log_{10} 95 = 1.977$ and $\log_{10} 2 = 0.3$)

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