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A string of 10 light bulbs is wired in such a way that if [#permalink]
 23 Apr 2012, 14:43
Question Stats:
48% (01:48) correct
51% (00:32) wrong based on 84 sessions
A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails. If for each individual light bulb the probability of failing during time period T is 0.06, what is the probability that the string of light bulbs will fail during the time period T? A. 0.06 B. (0.06)^10 C. 1 - (0.06)^10 D. (0.94)^10 E. 1 - (0.94)^10 I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ?
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
23 Apr 2012, 21:37
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Quote: I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be
0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ? It only takes one failure in any of the 10 slots to create the failure. For example if the second bulb fails, the whole string fails. If the 9th bulb fails, the string will fail too. Therefore the one and only chance for the strings to stay lit would be 10 consecutive non-failures. (.94)^10 = a lit string. Now it becomes 1-(.94)^10 since you're looking for the probability that the string of lights will fail. Hope that helps.
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
24 Apr 2012, 13:12
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massi2884 wrote: A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails. If for each individual light bulb the probability of failing during time period T is 0.06, what is the probability that the string of light bulbs will fail during the time period T?
A. 0.06
B. (0.06)^10
C. 1 - (0.06)^10
D. (0.94)^10
E. 1 - (0.94)^10
I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be
0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ? The string of light bulbs will fail if at least one light bulb fails. So, let's find the probability of the opposite event and subtract that value from 1. The opposite event is when none of the 10 light bulbs fails, since the probability of each light bulb not to fail is 1-0.06=0.94 the the probability that none of the 10 light bulbs fails is 0.94^10. Hence, the probability that at least one light bulb fails is 1-0.94^10. Answer: E. Now, you should have spotted that your reasoning was not right because of one simple thing, consider the case when we have 100 light bulbs instead of 10, then according to your logic the probability that the string of light bulbs will fail would be 100*0.06=6, which is not possible since the probability of an event can not be more than 1 (100%). Hope it's clear.
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
25 Apr 2012, 02:15
such questions can be answered as p(A) = 1 - P(not A)
Prob that bulb will fail is 0.06
hence, prob that bulb will light is 0.94
hence, prob that series will fail(i.e. at least one bulb will fail) = 1 -(All Bulbs light)
this is = 1-(0.94)^10
option E
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A string of 10 lightbulbs is wired in such a way that if any [#permalink]
16 May 2012, 03:41
A string of 10 lightbulbs is wired in such a way that if any individal lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during the period T is 0.06, what is the probability that the string of lightbulbs will fail during the period T?
A) 0.06
B) (0.06)^10
C) 1 - (0.06)^10
D) (0.94)^10
E) 1 - (0.94)^10
I still believe it is B. Why exactly is it E? I understand that E is 1- the probability of NOT failing. But I would assume the B is just the probability of it failing.
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Re: A string of 10 lightbulbs is wired in such a way that if any [#permalink]
16 May 2012, 03:47
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Merging similar topics. alexpavlos wrote: A string of 10 lightbulbs is wired in such a way that if any individal lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during the period T is 0.06, what is the probability that the string of lightbulbs will fail during the period T?
A) 0.06
B) (0.06)^10
C) 1 - (0.06)^10
D) (0.94)^10
E) 1 - (0.94)^10
I still believe it is B. Why exactly is it E? I understand that E is 1- the probability of NOT failing. But I would assume the B is just the probability of it failing. We are told that the string of light bulbs will fail if at least one light bulb fails. (0.06)^10 is the probability that all 10 lightbulbs will fail. Hope it's clear.
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COLLECTION OF QUESTIONS:
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Re: A string of 10 lightbulbs is wired in such a way that if any [#permalink]
16 May 2012, 04:45
Bunuel wrote: Merging similar topics. alexpavlos wrote: A string of 10 lightbulbs is wired in such a way that if any individal lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during the period T is 0.06, what is the probability that the string of lightbulbs will fail during the period T?
A) 0.06
B) (0.06)^10
C) 1 - (0.06)^10
D) (0.94)^10
E) 1 - (0.94)^10
I still believe it is B. Why exactly is it E? I understand that E is 1- the probability of NOT failing. But I would assume the B is just the probability of it failing. We are told that the string of light bulbs will fail if at least one light bulb fails. (0.06)^10 is the probability that all 10 lightbulbs will fail. Hope it's clear. Thank you its clear now! thanks again!!!
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
27 Nov 2012, 23:44
I understand why E is correct
but not understand why the result can not be 0.6. pls, explain using reasoning not using the quatitative.
why we can not add all possibilities?
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
17 Jan 2013, 02:20
thangvietnam wrote: I understand why E is correct
but not understand why the result can not be 0.6. pls, explain using reasoning not using the quatitative.
why we can not add all possibilities? SUGGESTED REASON: Since P(A or B) = P(A) + P(B) – P(AnB)....if 2 events can happen at the same time P(AnB) is not 0 There exists a probability that 2 or more lightbulbs can fail at the same time P(A or B or C) = P(A) + P(B) + P(C) – 2P(AnBnC) – [sum of exactly 2 groups members] So for 10 items, P(A or B or...or J) = P(A) + P(B)+...P(J) -...LOONG & COMPLEX calculations I hope my reasoning is correct?
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
24 Jan 2013, 01:37
hard question which appear after you get 48/51, I think.
very easy to understand E.
but why we can not add 0.06+0.06....=10x0.06
because
one possibility the system fail is that 0.06x(0.94)^9
we have 10 cases , so we add
10x 0.06 x (0.09)^9
I do not count. pls count for me. dose this way bring the result in E. pls help.
this way to help undersand the problem.
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A string of 10 light bulbs is wired in such [#permalink]
19 Feb 2013, 02:35
A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails.If for each individual light bulb the probability of falling during time period T is .06,what is the probability that the string of light bulbs will fail during during time period T?
(A).06
(B)(.06)^10
(C)1-(.06)^10
(D)(0.94)^10
(E)1-(0.94)^10
Need explanation.............
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Re: A string of 10 light bulbs is wired in such [#permalink]
19 Feb 2013, 03:22
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Re: A string of 10 light bulbs is wired in such
[#permalink]
19 Feb 2013, 03:22
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