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09-20-2008, 02:30 AM
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#1 | | Member
Join Date: May 2008 Location: Never-land
Posts: 677
| Difficult SAT word problem
A machine is made up of two components, A and B. Each component either works or fails. The failure or non-failure of one component is independent of the failure or non failure of the other component. The machine works if at least one of the components works. If the probability that each component works is 2/3, what is the probability that the machine works?
A. 4/9
B. 1/2
C. 2/3
D. 5/6
E. 8/9
Please explain your answer, thanks! | 
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09-20-2008, 02:39 AM
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#2 | | Senior Member
Join Date: Jul 2008
Posts: 1,179
| Is the answer E?
I guess I'll to explain if I'm correct... | |
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09-20-2008, 02:41 AM
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#3 | | Member
Join Date: May 2008 Location: Never-land
Posts: 677
| You are correct. I understand you can do this by finding the probability of the machine not working, which is 1/3. 1/3 x 1/3 = 1/9 1-1/9 = 8/9. However, what i dont get is why we cant do it vice versa. why can't we do 2/3 x 2/3 ? | |
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09-20-2008, 02:44 AM
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#4 | | Senior Member
Join Date: Jul 2008
Posts: 1,179
| Hmm, (2/3)x(2/3) would give you the probability that BOTH components work; however, only one of the components need to be working for the machine to work. Eh, there's probably another way to do this, but I just did it by figuring out the probability that the machine would not work, which would be both components not working (1/3)x(1/3) and subtracting that from 1 to yield the probability that the machine would work. | |
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09-20-2008, 02:45 AM
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#5 | | Senior Member
Join Date: Jun 2007 Location: Honolulu -> ??? '13
Posts: 3,271
| The only time the machine doesn't work is when both components don't work (1/3 for both components). Multiple 1/3X1/3 for the one time when both don't work and the machine doesn't work. Subtract that from 1. 8/9. | |
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09-20-2008, 02:46 AM
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#6 | | Senior Member
Join Date: Jul 2008
Posts: 1,179
| I love how people continue to post the same solution even after the problem has been solved lol. | |
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09-20-2008, 03:07 AM
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#7 | | Senior Member
Join Date: May 2008 Location: Guam
Posts: 1,505
| O.o took me a while,i got E as well | 
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09-20-2008, 12:08 PM
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#8 | | Senior Member
Join Date: Mar 2008 Location: Out of my mind. Back in five minutes.
Posts: 2,504
| Quote: |
I love how people continue to post the same solution even after the problem has been solved lol.
| its because they don't want to the the one that's "wrong" lol  | |
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09-20-2008, 04:39 PM
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#9 | | Member
Join Date: Jun 2008
Posts: 575
| Well I guess your logic comes into this too. If you know whether the probability is over half or not, then you should be able to catch ur own mistake if you pick 4/9. | |
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09-20-2008, 04:56 PM
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#10 | | Member
Join Date: Jun 2008
Posts: 624
| The chance that 2 works is 4/9. The probability that only A works but B doesnt is 2/3 * 1/3 = 2/9. The probability that B works and A doesn't is 2/3 * 1/3 = 2/9.
Therefore the total is 4/9+2/9+2/9 = 8/9.
The other way is simpler though, so no point doing it this way. | |
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09-20-2008, 05:12 PM
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#11 | | Member
Join Date: Jun 2008
Posts: 575
| oh wow. that really put it into perspective khotrinh, thanks. | |
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09-20-2008, 06:00 PM
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#12 | | Senior Member
Join Date: Jun 2007 Location: Honolulu -> ??? '13
Posts: 3,271
| I love how people continue to post the same solution even after the problem has been solved lol.
It wasn't solved when I originally opened the topic. I open like 10 webpages at a time so it took me a little while to get to this one and answer the question. By the time I posted I saw that it was already solved but I didn't care enough to edit my post.
I love how people can't look at the post times and make simple logical judgments. | |
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09-20-2008, 06:07 PM
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#13 | | New Member
Join Date: Sep 2008
Posts: 22
| here's a graphical way to represent it, for anybody that wants...
i figured i would help because people are usually pretty helpful here  http://localhostr.com/files/806500/answer.png | |
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09-20-2008, 10:17 PM
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#14 | | Junior Member
Join Date: Mar 2008
Posts: 223
| That's a pretty impressive diagram andrea giovanni. I think you waste too much time on an exam drawing that, but it's a nice way to look at the problem. | |
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