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A leather pouch contains 10 light bulbs, two of which are de
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29 Oct 2017, 01:18
29 Oct 2017, 01:18
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65% (00:51) correct
34% (01:03) wrong based on 127 sessions
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A leather pouch contains 10 light bulbs, two of which are defective.
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Kudos for correct solution.
Quantity A |
Quantity B |
The probability of drawing two defective light bulbs (without replacement) |
The probability of drawing two defective light bulbs (without replacement) and then a non-defective bulb |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Kudos for correct solution.
Re: A leather pouch contains 10 light bulbs, two of which are de
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30 Oct 2017, 07:01
30 Oct 2017, 07:01
4
The probability of extracting two defective light bulbs without replacement is equal to \(\frac{2}{10}*\frac{1}{9} = \frac{1}{45}\).
The probability of extracting two defective light bulbs without replacement and then a non-defective one is equal to \(\frac{2}{10}*\frac{1}{9}*\frac{8}{8} = \frac{1}{45}\).
The idea is that there are 2 defective light bulbs out of 10 in the first draw, then there is just one out of the 9 light bulbs remained. Once the two defective light bulbs are extracted there are eight out of 8 non defective light bulbs.
The two quantities are equal.
Answer C
The probability of extracting two defective light bulbs without replacement and then a non-defective one is equal to \(\frac{2}{10}*\frac{1}{9}*\frac{8}{8} = \frac{1}{45}\).
The idea is that there are 2 defective light bulbs out of 10 in the first draw, then there is just one out of the 9 light bulbs remained. Once the two defective light bulbs are extracted there are eight out of 8 non defective light bulbs.
The two quantities are equal.
Answer C
Re: A leather pouch contains 10 light bulbs, two of which are de
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20 Dec 2018, 14:54
20 Dec 2018, 14:54
Answer: C
10 light bulbs: 8 non-defective + 2 detective
A: The probability of drawing two defective light bulbs (without replacement):
For taking the first one, the probability of being defective is 2/10, for the next one the probability is 1/9, so A equals: 2/10 * 1/ 9 = 2 / 90
B: The probability of drawing two defective light bulbs (without replacement) and then a non-defective bulb:
For taking the first one, the probability of being defective is 2/10, for the next one the probability is 1/9, for the third one, because now we have already taken the two defective lights and there are 8 non-defective lights, the probability of taking a non-defective light is 8/8, so B equals: 2/10 * 1/ 9 * 8/8 = 2 / 90
A and B are equal. The answer is C.
10 light bulbs: 8 non-defective + 2 detective
A: The probability of drawing two defective light bulbs (without replacement):
For taking the first one, the probability of being defective is 2/10, for the next one the probability is 1/9, so A equals: 2/10 * 1/ 9 = 2 / 90
B: The probability of drawing two defective light bulbs (without replacement) and then a non-defective bulb:
For taking the first one, the probability of being defective is 2/10, for the next one the probability is 1/9, for the third one, because now we have already taken the two defective lights and there are 8 non-defective lights, the probability of taking a non-defective light is 8/8, so B equals: 2/10 * 1/ 9 * 8/8 = 2 / 90
A and B are equal. The answer is C.
