GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
A leather pouch contains 10 light bulbs, two of which are de
[#permalink]
29 Oct 2017, 01:18
29 Oct 2017, 01:18
2
6
Bookmarks
00:00
Question Stats:
66% (00:51) correct
33% (01:06) wrong based on 136 sessions
Hide Show timer Statistics
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
[#permalink]
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
[#permalink]
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.
