Question:
Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is
Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is
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Drawing without replacement follows Hypergeometric distribution.
Updated On: Apr 26, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Total and Defective
Total oranges = 20. Defective (D) = 4, Good (G) = 16. Sample size = 3.
Step 2: Probabilities
- $P(0) = \frac{16}{20} \cdot \frac{15}{19} \cdot \frac{14}{18} = \frac{4}{5} \cdot \frac{15}{19} \cdot \frac{7}{9} = \frac{28}{57}$.
- $P(3) = \frac{4}{20} \cdot \frac{3}{19} \cdot \frac{2}{18} = \frac{1}{5} \cdot \frac{3}{19} \cdot \frac{1}{9} = \frac{1}{285}$.
Step 3: Conclusion
Based on hypergeometric calculations, the values in the table for option A or B in the paper match these probabilities.
Final Answer: (A)
Total oranges = 20. Defective (D) = 4, Good (G) = 16. Sample size = 3.
Step 2: Probabilities
- $P(0) = \frac{16}{20} \cdot \frac{15}{19} \cdot \frac{14}{18} = \frac{4}{5} \cdot \frac{15}{19} \cdot \frac{7}{9} = \frac{28}{57}$.
- $P(3) = \frac{4}{20} \cdot \frac{3}{19} \cdot \frac{2}{18} = \frac{1}{5} \cdot \frac{3}{19} \cdot \frac{1}{9} = \frac{1}{285}$.
Step 3: Conclusion
Based on hypergeometric calculations, the values in the table for option A or B in the paper match these probabilities.
Final Answer: (A)
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