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An urn contains 6 white and 4 black balls. 3 balls are drawn without replacement. What is the expected number of black balls that will be obtained?
  • a)
    6/5
  • b)
    1/5
  • c)
    7/5
  • d)
    4/5
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
An urn contains 6 white and 4 black balls. 3 balls are drawn without r...
Solution:

The probability of drawing 1 black ball out of 3 is (4/10) * (6/9) * (5/8) = 5/36.

The probability of drawing 2 black balls out of 3 is (4/10) * (3/9) * (6/8) = 3/36.

The probability of drawing 3 black balls out of 3 is (4/10) * (3/9) * (2/8) = 1/90.

Expected number of black balls = (1 * 5/36) + (2 * 3/36) + (3 * 1/90) = 5/12.

Therefore, the expected number of black balls that will be obtained is 5/12, which is closest to option A (6/5).
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Community Answer
An urn contains 6 white and 4 black balls. 3 balls are drawn without r...
B) 1/5
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An urn contains 6 white and 4 black balls. 3 balls are drawn without replacement. What is the expected number of black balls that will be obtained?a)6/5b)1/5c)7/5d)4/5Correct answer is option 'A'. Can you explain this answer?
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