select the correct answer from each drop-down menu.

a container holds 50 electronic components, of which 10 are defective.

if 6 components are drawn at random from the container, the probability that at least 4 are not defective is choose an answer. if 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is choose an answer.

Question

select the correct answer from each drop-down menu.

a container holds 50 electronic components, of which 10 are defective.

if 6 components are drawn at random from the container, the probability that at least 4 are not defective is choose an answer. if 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is choose an answer.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Hypergeometric Probability", "Combinations" ], "new_concepts": [], "current_concepts": [ "Hypergeometric Probability", "Combinations", "Classical Probability" ] } ## Calculate probability of drawing at least 4 non-defective components out of 6 $$ \begin{aligned} &N = 50,\quad K_{\text{not defective}} = 40,\quad K_{\text{defective}} = 10,\quad n = 6\ &P(X \ge 4) = P(X = 4) + P(X = 5) + P(X = 6)\ &P(X = 4) = \frac{\binom{40}{4}\binom{10}{2}}{\binom{50}{6}} = \frac{91390 \times 45}{15890700} \approx 0.2588\ &P(X = 5) = \frac{\binom{40}{5}\binom{10}{1}}{\binom{50}{6}} = \frac{658008 \times 10}{15890700} \approx 0.4141\ &P(X = 6) = \frac{\binom{40}{6}\binom{10}{0}}{\binom{50}{6}} = \frac{3838380 \times 1}{15890700} \approx 0.2415\ &P(X \ge 4) \approx 0.2588 + 0.4141 + 0.2415 = 0.9144 \approx 0.914 \end{aligned} $$ ## Calculate probability of drawing exactly 3 defective components out of 8 $$ \begin{aligned} &N = 50,\quad K_{\text{defective}} = 10,\quad K_{\text{not defective}} = 40,\quad n = 8\ &P(Y = 3) = \frac{\binom{10}{3}\binom{40}{5}}{\binom{50}{8}}\ &\binom{10}{3} = 120\ &\binom{40}{5} = 658008\ &\binom{50}{8} = 536878650\ &P(Y = 3) = \frac{120 \times 658008}{536878650} = \frac{78960960}{536878650} \approx 0.147 \end{aligned} $$ Select the correct answer from each drop-down menu. A container holds 50 electronic components, of which 10 are defective. If 6 components are drawn at random from the container, the probability that at least 4 are *not* defective is 0.914. If 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is 0.147. { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Hypergeometric Probability" ] }

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Question

Explanation:

Calculate probability of drawing at least 4 non-defective components out of 6

Calculate probability of drawing exactly 3 defective components out of 8

Answer: