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12 Jun 2018, 11:11
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In a retail shop, 32% pens are defective, and both defective as well as non-defective pens are identical and kept together so there is no way to tell defective from non-defective. If a customer buys 3 pens randomly, what is the probability that none of the 3 pens will be defective?
(1) If one pen is chosen randomly, the probability of that pen being non-defective is 17/25.
(2) There are total 50 pens in the shop.
(1) If one pen is chosen randomly, the probability of that pen being non-defective is 17/25.
(2) There are total 50 pens in the shop.
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12 Jun 2018, 12:01
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In a retail shop, 32% pens are defective, and both defective as well as non-defective pens are identical and kept together so there is no way to tell defective from non-defective. If a customer buys 3 pens randomly, what is the probability that none of the 3 pens will be defective?
(1) If one pen is chosen randomly, the probability of that pen being non-defective is 17/25.
(2) There are total 50 pens in the shop.
>> (2): Sufficient
Total = 50 pens
0.32*50 = 16 defectives
34 Non defectives
Ways of choosing 3 non defective out of 34 fine pieces of pen = 34C3
Ways of choosing 3 out of 50 pieces of pen = 50C3
>>> 34C3 / 50C3 = 374/1225
>> (1): Insufficient
17/25 = 68/100 (hence, 68% are non defective. we know that from the statement)
No information about the total number of pens.
Option B.
(1) If one pen is chosen randomly, the probability of that pen being non-defective is 17/25.
(2) There are total 50 pens in the shop.
>> (2): Sufficient
Total = 50 pens
0.32*50 = 16 defectives
34 Non defectives
Ways of choosing 3 non defective out of 34 fine pieces of pen = 34C3
Ways of choosing 3 out of 50 pieces of pen = 50C3
>>> 34C3 / 50C3 = 374/1225
>> (1): Insufficient
17/25 = 68/100 (hence, 68% are non defective. we know that from the statement)
No information about the total number of pens.
Option B.
21 Jun 2018, 01:58
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just a question. How is the value of total number of pens important for finding out the probability and even percentage can give us the probability.
For me, both the options are sufficient.
Please correct if i am wrong in my concept.
For me, both the options are sufficient.
Please correct if i am wrong in my concept.
