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A chain is comprised of 10 identical links, each of which in [#permalink]
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05 Jan 2013, 06:35
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A chain is comprised of 10 identical links, each of which independently has a 1% chance of breaking under a certain load. If the failure of any individual link means the failure of the entire chain, what is the probability that the chain will fail under the load? (A) \((0.01)^{10}\) (B) \(10(0.01)^{10}\) (C) \(1(0.10)(0.99)^{10}\) (D) \(1(0.99)^{10}\) (E) \(1(0.99)^{(10*9)}\)
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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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05 Jan 2013, 07:01
Optimus66 wrote: A chain is comprised of 10 identical links, each of which independently has a 1% chance of breaking under a certain load. If the failure of any individual link means the failure of the entire chain, what is the probability that the chain will fail under the load?
(A) \((0.01)^{10}\) (B) \(10(0.01)^{10}\) (C) \(1(0.10)(0.99)^{10}\) (D) \(1(0.99)^{10}\) (E) \(1(0.99)^{(10*9)}\) There are 10 links, each of which has the probability of 0.99 of not failing. If any of them breaks, the chain will not survive. To find the probability that the chain will fail under the load implies to find the the probability that any link breaks. To do so, find the reverse probability: the probability that the chain will not break or none of the 10 links break. Since there are 10 links, hence probability that none of the link breaks is \(0.99^{10}\). Hence P that the chain will break is \(10.99^{10}\).
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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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05 Jan 2013, 12:12
Background: The whole chain will fail if atleast one of the 10 links fails. Probability [that a single link will NOT break] \(= 0.99\) Probability [that none of the 10 links will break] \(= (0.99)^{10}\) Thus probability [that atleast 1 of the link will break] \(= 1  (0.99)^{10}\) Hence (D).
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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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07 Jan 2013, 01:47



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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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08 Jan 2013, 12:02
Marcab wrote: Optimus66 wrote: A chain is comprised of 10 identical links, each of which independently has a 1% chance of breaking under a certain load. If the failure of any individual link means the failure of the entire chain, what is the probability that the chain will fail under the load?
(A) \((0.01)^{10}\) (B) \(10(0.01)^{10}\) (C) \(1(0.10)(0.99)^{10}\) (D) \(1(0.99)^{10}\) (E) \(1(0.99)^{(10*9)}\) There are 10 links, each of which has the probability of 0.99 of failing. If any of them breaks, the chain will not survive. To find the probability that the chain will fail under the load implies to find the the probability that any link breaks. To do so, find the reverse probability: the probability that the chain will not break or none of the 10 links break. Since there are 10 links, hence probability that none of the link breaks is \(0.99^{10}\). Hence P that the chain will break is \(10.99^{10}\). not failing you mean! am sure you got caught up!
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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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08 Jan 2013, 19:18



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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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16 May 2013, 14:56
My three weaknesses on the GMAT are combinatorics, word problems, and probability!!!
I am having difficulty figuring out how to solve this problem. The book says to take the probability of each link NOT failing, multiplying them by one another and subtracting the result from one. I understand superficially that this is a time/work saving measure, but why does it work and how would I know how to do that come test time?!
While we're on the topic, does anyone have a good guide on basic, intermediate and advanced probability/combinatoric skills? I have been using the Manhattan guides and in general they are quite good, but they don't have a lot of problems on probability nor do they have many lessons lessons on them.
Thanks!



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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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16 May 2013, 23:38
WholeLottaLove wrote: My three weaknesses on the GMAT are combinatorics, word problems, and probability!!!
I am having difficulty figuring out how to solve this problem. The book says to take the probability of each link NOT failing, multiplying them by one another and subtracting the result from one. I understand superficially that this is a time/work saving measure, but why does it work and how would I know how to do that come test time?!
While we're on the topic, does anyone have a good guide on basic, intermediate and advanced probability/combinatoric skills? I have been using the Manhattan guides and in general they are quite good, but they don't have a lot of problems on probability nor do they have many lessons lessons on them.
Thanks! This topic might help with your doubts: astringof10lightbulbsiswiredinsuchawaythatif131205.htmlIn addition check probability chapter of Math Book for theory: mathprobability87244.htmlAlso check some probability questions to practice: DS: search.php?search_id=tag&tag_id=33PS: search.php?search_id=tag&tag_id=54Hard questions on combinations and probability with detailed solutions: hardestareaquestionsprobabilityandcombinations101361.html (there are some about permutation too) Hope it helps.
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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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17 Aug 2015, 23:47
daviesj wrote: A chain is comprised of 10 identical links, each of which independently has a 1% chance of breaking under a certain load. If the failure of any individual link means the failure of the entire chain, what is the probability that the chain will fail under the load?
(A) \((0.01)^{10}\) (B) \(10(0.01)^{10}\) (C) \(1(0.10)(0.99)^{10}\) (D) \(1(0.99)^{10}\) (E) \(1(0.99)^{(10*9)}\) It is easier to find if you apply 1 P(the chain never breaks)=P( the chain will fail) the probability that one link does not break is 1.01=.99 the entire chain does not break if the first link does not break and the second link does not break and the third link does not break and so on till the tenth link therefore, 0.99*0.99*0.99*... ten times so, the answer is 1(0.99)^10 Hence, the correct option is D



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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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12 Nov 2016, 12:20
daviesj wrote: A chain is comprised of 10 identical links, each of which independently has a 1% chance of breaking under a certain load.If the failure of any individual link means the failure of the entire chain, what is the probability that the chain will fail under the load?
(A) \((0.01)^{10}\) (B) \(10(0.01)^{10}\) (C) \(1(0.10)(0.99)^{10}\) (D) \(1(0.99)^{10}\) (E) \(1(0.99)^{(10*9)}\) Probability of failure = Probability of non failure of atleast 1 link Probability of failure of the chain = 1  Probability of no failure Probability of failure of the chain = \(1(0.99)^{10}\) Hence, answer will definitely be (D) \(1(0.99)^{10}\)
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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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14 Mar 2017, 17:09
Qualitatively, many failure scenarios could occur: • None of the links will fail, • Exactly 1 of the links will fail, • Exactly 2 of the links will fail, • etc. Given the complexity of the failure scenarios, it is easier for us to look at the opposite scenario: probability that at least 1 link will fail = 1 – probability that all links will not fail For each of the links, the probability that it will not fail is 1 – 0.01 = 0.99. The probability that all ten will not fail is thus (0.99)^10, since the probability that all ten will not fail is simply the product of the probabilities of the individual links not failing. Therefore, the Probability that at least 1 link will fail = 1 – (0.99)^10. The correct answer is D
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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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20 Mar 2017, 06:50
daviesj wrote: A chain is comprised of 10 identical links, each of which independently has a 1% chance of breaking under a certain load. If the failure of any individual link means the failure of the entire chain, what is the probability that the chain will fail under the load?
(A) \((0.01)^{10}\) (B) \(10(0.01)^{10}\) (C) \(1(0.10)(0.99)^{10}\) (D) \(1(0.99)^{10}\) (E) \(1(0.99)^{(10*9)}\) We can use the following equation: 1 = P(at least 1 of 10 links failing) + P(nolinks failing) Since we need to determine the probability that the chain will fail, we are looking for P(at least 1 of 10 links failing). However, we can solve for P(nolinks failing) and then subtract that probability from 1. Since each link has a 1% chance of failing, each link has a 99% chance of NOT failing. Thus, P(nolinks failing) = (0.99)^10, so: P(at least 1 of 10 links failing) = 1  (0.99)^10. Answer: D
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Re: A chain is comprised of 10 identical links, each of which in [#permalink]
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21 Mar 2017, 04:20
daviesj wrote: A chain is comprised of 10 identical links, each of which independently has a 1% chance of breaking under a certain load. If the failure of any individual link means the failure of the entire chain, what is the probability that the chain will fail under the load?
(A) \((0.01)^{10}\) (B) \(10(0.01)^{10}\) (C) \(1(0.10)(0.99)^{10}\) (D) \(1(0.99)^{10}\) (E) \(1(0.99)^{(10*9)}\) P(link fails) = 1  P(link not fail) = 1  (10.01)(10.01)(10.01)(10.01)(10.01)(10.01)(10.01)(10.01)(10.01)(10.01) = 1  0.99^10




Re: A chain is comprised of 10 identical links, each of which in
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