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A string of 10 light bulbs is wired in such a way that if [#permalink]
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23 Apr 2012, 14:43
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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]
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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 nonfailures. (.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]
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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 10.06=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 10.94^10. 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 cannot 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]
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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]
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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]
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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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Re: A string of 10 lightbulbs is wired in such a way that if any [#permalink]
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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]
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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]
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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]
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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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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
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16 Jan 2014, 23:28
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 ? Probability of Failing = 1  probability of success Probability of failing = 1  (1  0.06)^10 = 1  (0.94)^10
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
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17 Jan 2014, 06:37
Can please someone explain why answer A is wrong? We are told that the entire wire will fail if one bulb fails. Since the probability of one bulb to fail is 0.6 then the probability of the entire wire to fail should be also 0.6.



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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
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12 May 2015, 00:41
Bunuel wrote: Dmba wrote: Can please someone explain why answer A is wrong? We are told that the entire wire will fail if one bulb fails. Since the probability of one bulb to fail is 0.6 then the probability of the entire wire to fail should be also 0.6. We are told that the string of light bulbs will fail if at least one light bulb fails. So, the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, 1 minus the probability that neither fails. Complete solution is here: astringof10lightbulbsiswiredinsuchawaythatif131205.htmlHope it helps. Bunuel, can you solve this question with alternative method that you mentioned above  the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, [i]1 minus the probability that neither fails Did not give the same result.
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
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Ergenekon wrote: Bunuel wrote: Dmba wrote: Can please someone explain why answer A is wrong? We are told that the entire wire will fail if one bulb fails. Since the probability of one bulb to fail is 0.6 then the probability of the entire wire to fail should be also 0.6. We are told that the string of light bulbs will fail if at least one light bulb fails. So, the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, 1 minus the probability that neither fails. Complete solution is here: astringof10lightbulbsiswiredinsuchawaythatif131205.htmlHope it helps. Bunuel, can you solve this question with alternative method that you mentioned above  the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, [i]1 minus the probability that neither fails Did not give the same result. Hi Ergenekon, Let me write the probability equation for event method:P(String of lights failing) = P(1 or 2 or .....10 lights failing) P(String of lights failing) = P(1 Fail & 9 Lit) + P(2 Fail & 8 Lit) +..........P( 10 Fail & 0 Lit) P(String of lights failing) = \({10_C_1}\) \(* 0.06^1*0.94^9 +\) \({10_C_2}\) \(* 0.06^2 *0.94^8 +\).............\({10_C_{10}}\) \(* 0.06^{10}*0.94^0\) The important point here is to select the bulbs which would fail out of the total bulbs for each case. So, when we write the probability equation for 1 bulb failing, there can be 10 ways in which a bulb can fail. Similarly, when you write the probability equation for the nonevent method i.e. P(String of lights failing) = 1  P(All lights lit) \(= 1 \) \({10_C_{10}}\) \(* 0.94^{10}\) Both the probability equations will give you the same result. However you would notice that the nonevent method is far more easier to calculate and comprehend than the event method. For a probability question it is recommended to evaluate the number of cases in the event and the nonevent method before proceeding with the solution. Hope this helps Regards Harsh
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
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15 May 2016, 08:41
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
Probability of NOT failing one light bulb = 1.06= 0.94 Probability of NOT failing all light bulbs = 0.94^10 Probability of failing= 10.94^10 E is the answer
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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
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25 Jun 2016, 21:29
For those not understanding this questing and wondering why you cannot do (6/100)^10
here is why
Let's make a simple example with using 2 lights and them each having a probability of failure of 1/10, and the chain fails if atleast 1 light fails.
What is the probability of failure?
Quick way: 1 P(success) > 1  (9/10)^2 = 19/100
Other Way
So why cant we do (1/10)^2 ... well we CAN we are just considering 1 case in which the strand fails. It also fails with light 1 success and light 2 fail
so we have
P(fail)*P(fail) > 1/100 + P(success)*P(fail) > 9/100 + P(fail)*P(success) > 9/100
Together we get the probability of failure = 19/100
SO back to the 700 level question, the long way would be EXTREMELY cumbersome, so keep the 1p(success) method in your mind ! The long way may only work for a 600 level question!



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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
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26 Jun 2016, 00:30
It's important to realize that the question is asking for the probability that at least one light bulb will fail. When a question asks for the probability of "at least one", the easiest approach is to calculate the probability of "none" and take the complement of it. So the answer is 1  (10.06)^10. Answer choice B.
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