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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) 6/321 and 3/926 (b) 1/20 and 1/30 (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 ba...
Solution:
With Replacement
Total number of balls = 12
Probability of drawing a white ball on the first draw = 5/12
Probability of drawing a white ball on the second draw = 5/12
Probability of drawing a black ball on the third draw = 7/12
Probability of drawing 3 balls with replacement = (5/12) * (5/12) * (7/12) = 175/10368
Probability of the first draw producing white balls and the second draw producing black balls = 175/10368
Without Replacement
Total number of balls = 12
Probability of drawing a white ball on the first draw = 5/12
Probability of drawing a white ball on the second draw = 4/11
Probability of drawing a black ball on the third draw = 7/10
Probability of drawing 3 balls without replacement = (5/12) * (4/11) * (7/10) = 7/132
Probability of the first draw producing white balls and the second draw producing black balls = 7/132
Therefore, the correct option is (d) 7/968 and 5/264.
Explanation:
The problem can be solved by using the formula for conditional probability. The probability of the first event (drawing a white ball on the first draw) is multiplied by the probability of the second event (drawing a white ball on the second draw given that a white ball was drawn on the first draw) and then multiplied by the probability of the third event (drawing a black ball on the third draw given that white balls were drawn on the first two draws).
The difference between the two scenarios (with replacement and without replacement) is that in the first scenario, the probability of drawing a white ball on the second draw is the same as the probability of drawing a white ball on the first draw, while in the second scenario, the probability of drawing a white ball on the second draw is different (it is one less than the number of white balls remaining in the box divided by the total number of balls remaining in the box).
Therefore, the probability of the first draw producing white balls and the second draw producing black balls is higher in the scenario with replacement than in the scenario without replacement.
With Replacement
Total number of balls = 12
Probability of drawing a white ball on the first draw = 5/12
Probability of drawing a white ball on the second draw = 5/12
Probability of drawing a black ball on the third draw = 7/12
Probability of drawing 3 balls with replacement = (5/12) * (5/12) * (7/12) = 175/10368
Probability of the first draw producing white balls and the second draw producing black balls = 175/10368
Without Replacement
Total number of balls = 12
Probability of drawing a white ball on the first draw = 5/12
Probability of drawing a white ball on the second draw = 4/11
Probability of drawing a black ball on the third draw = 7/10
Probability of drawing 3 balls without replacement = (5/12) * (4/11) * (7/10) = 7/132
Probability of the first draw producing white balls and the second draw producing black balls = 7/132
Therefore, the correct option is (d) 7/968 and 5/264.
Explanation:
The problem can be solved by using the formula for conditional probability. The probability of the first event (drawing a white ball on the first draw) is multiplied by the probability of the second event (drawing a white ball on the second draw given that a white ball was drawn on the first draw) and then multiplied by the probability of the third event (drawing a black ball on the third draw given that white balls were drawn on the first two draws).
The difference between the two scenarios (with replacement and without replacement) is that in the first scenario, the probability of drawing a white ball on the second draw is the same as the probability of drawing a white ball on the first draw, while in the second scenario, the probability of drawing a white ball on the second draw is different (it is one less than the number of white balls remaining in the box divided by the total number of balls remaining in the box).
Therefore, the probability of the first draw producing white balls and the second draw producing black balls is higher in the scenario with replacement than in the scenario without replacement.
Community Answer
A box contains 5 white and 7 black balls. Two successive drawn of 3 ba...
Option c
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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) 6/321 and 3/926 (b) 1/20 and 1/30 (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) 6/321 and 3/926 (b) 1/20 and 1/30 (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) 6/321 and 3/926 (b) 1/20 and 1/30 (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) 6/321 and 3/926 (b) 1/20 and 1/30 (c) 35/144 and 35/108 (d) 7/968 and 5/264?.
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) 6/321 and 3/926 (b) 1/20 and 1/30 (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) 6/321 and 3/926 (b) 1/20 and 1/30 (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) 6/321 and 3/926 (b) 1/20 and 1/30 (c) 35/144 and 35/108 (d) 7/968 and 5/264?.
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