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Updated on: 25 Jun 2015, 06:14
Originally posted by kanigmat011 on 25 Jun 2015, 06:10.
Last edited by Bunuel on 25 Jun 2015, 06:14, edited 4 times in total.
Last edited by Bunuel on 25 Jun 2015, 06:14, edited 4 times in total.
Renamed the topic and edited the question.
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (02:30) correct
31%
(02:31)
wrong
based on 296
sessions
History
Date
Time
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Not Attempted Yet
A box contains 20 bulbs, 12 are defective. Four bulbs are selected at random. What is the probability that at least one of the four selected is defective?
A. 33/323
B. 1/9690
C. 14/969
D. 955/969
E. 5/12
A. 33/323
B. 1/9690
C. 14/969
D. 955/969
E. 5/12
25 Jun 2015, 06:19
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kanigmat011
12 are defective, 8 not defective.
P(at least one defective out of 4) = 1 - P(none is defective out of 4) = 1 - 8/20*7/19*6/18*5/17 = 955/969.
Answer: D.
25 Jun 2015, 06:34
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kanigmat011
Bunuel's solution is perfect however another insight to answer the questions is
Since 60% are defective (12/20) therefore the probability of one of the 4 selected bulb defective should be pretty high and should be close to 1
only Option D gives a close enough value as in every other option the probability is even less than 1/2
Answer: Option D
