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A box contains 5 white and 7 black balls. Two successive drawn of 3 balls are made (i) with replacement (ii) without replacement. The probability that the first draw would produce white balls and the second draw would produce black balls are respectively
(a) 1/20 and 1/30
(b) 6/321 and 3/926
(c) 35/144 and 35/108
(d) 7/968 and 5/264?
Most Upvoted Answer
A box contains 5 white and 7 black balls. Two successive drawn of 3 ba...
Understanding the Problem
In this problem, we need to calculate the probabilities of drawing white and black balls under two different conditions: with replacement and without replacement.
Case (i): With Replacement
- Total balls: 5 white + 7 black = 12 balls.
- Probability of drawing 3 white balls:
- Probability of drawing 1 white ball = 5/12.
- Probability of drawing 3 white balls = (5/12) * (5/12) * (5/12) = 125/1728.
- Probability of drawing 3 black balls:
- Probability of drawing 1 black ball = 7/12.
- Probability of drawing 3 black balls = (7/12) * (7/12) * (7/12) = 343/1728.
- Now, the probability of the first draw yielding white balls and the second yielding black balls:
- P(White, then Black) = P(3 White) * P(3 Black) = (125/1728) * (343/1728) = 42875/2985984 ≈ 1/20.
Case (ii): Without Replacement
- The total number of ways to draw 3 balls from 12 is C(12,3) = 220.
- Probability of drawing 3 white balls:
- C(5,3) = 10.
- P(3 White) = 10/220 = 1/22.
- Probability of drawing 3 black balls after 3 white:
- Remaining balls: 2 white + 7 black = 9.
- C(7,3) = 35.
- P(3 Black) = 35/84 = 5/12.
- Thus, P(White, then Black) = P(3 White) * P(3 Black) = (1/22) * (5/12) = 5/264.
Final Results
- The probabilities for the two cases are:
- With Replacement: 1/20.
- Without Replacement: 5/264.
Hence, the correct answer is option (a) 1/20 and 5/264.
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A box contains 5 white and 7 black balls. Two successive drawn of 3 balls are made (i) with replacement (ii) without replacement. The probability that the first draw would produce white balls and the second draw would produce black balls are respectively (a) 1/20 and 1/30 (b) 6/321 and 3/926 (c) 35/144 and 35/108 (d) 7/968 and 5/264?
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A box contains 5 white and 7 black balls. Two successive drawn of 3 balls are made (i) with replacement (ii) without replacement. The probability that the first draw would produce white balls and the second draw would produce black balls are respectively (a) 1/20 and 1/30 (b) 6/321 and 3/926 (c) 35/144 and 35/108 (d) 7/968 and 5/264? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about A box contains 5 white and 7 black balls. Two successive drawn of 3 balls are made (i) with replacement (ii) without replacement. The probability that the first draw would produce white balls and the second draw would produce black balls are respectively (a) 1/20 and 1/30 (b) 6/321 and 3/926 (c) 35/144 and 35/108 (d) 7/968 and 5/264? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A box contains 5 white and 7 black balls. Two successive drawn of 3 balls are made (i) with replacement (ii) without replacement. The probability that the first draw would produce white balls and the second draw would produce black balls are respectively (a) 1/20 and 1/30 (b) 6/321 and 3/926 (c) 35/144 and 35/108 (d) 7/968 and 5/264?.
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