<p>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? </p>
<p>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 ?</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>Therefore the total is 4/9+2/9+2/9 = 8/9.</p>
<p>The other way is simpler though, so no point doing it this way.</p>
<p>I love how people continue to post the same solution even after the problem has been solved lol.</p>
<p>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. </p>
<p>I love how people can't look at the post times and make simple logical judgments.</p>
<p>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.</p>