<p>Ok. so this isn't sat math, but I really need help with this problem. Hopefully you guys can explain the process not just the answer. here it is:
Two machines, A and B each have 3 components that are required for production to continue. Machine A will function only if all three components work. On the other hand, machine B can function if one of it components is working. Each component works independently of the others and fails with probability p. If machine A breaks down 20% more than machine B , find p.</p>
<p>Ok so this is what I thought about doing:
P(machine B failing)= (p)^3 ( because all three components have to be failing for it to stop working)</p>
<p>P (machine A failing)= 1- (1-p^3)...i thought 1. the probablility of machine a wokring ( 3 components working) is the probablity that it'd fail</p>
<p>and i guess then P(MA)= P(MB)+0.2</p>
<p>so</p>
<p>p^3+0.2= 1-(1-p^3)... is this even ok? HELP please</p>
<p>Also, the phrase '...breaks down 20% more than machine B' is a bit ambiguous. The way you construed it is probably correct i.e.
1 -(1-p)^3 = 0.2 + p^3 ----------------- (1)</p>
<p>but (IMO), someone could also read it as
1 - (1-p)^3 = 1.2 (p^3) ----------------- (2)</p>
<p>However, eq.(2) leaves you with a cubic equation to solve for p, whereas eq. (1) leaves you with a quadratic. (2) is unlikely - go with (1). If you expand both sides fully, the p^3 will cancel on both sides, leaving you with a quadratic equation in p.</p>
<p>ok, awesom thanks... Oh btw I had been meaning to ask you. Recently I've decided that I might end up studying econ. and some sort of engineering. My real passion is math, along with econ, but my parents refuse to pay for my education if I decided to study math, so I chose what they suggested as a close alternative: engineering. So i was jsut wondering, how in depth does an engineer get to study math? It might sound like a dumb question but nobody decided to do engineering in my family so I have had little to no contact with this career.</p>
<p>I have an undergrad degree in electronics engineering, but from India, not the US. Went through more freaking math than I <em>ever</em> wanted :) .</p>
<p>To some extent, it depends on the branch of engineering you choose. But I think it's safe to say that you'll find all the math you ever wanted. Hey, back in my day engineers used to walk around with slide-rules strapped to their belts, like six-shooters. Had favorite logarithms memorized, and square roots...</p>
<p>hehe, sounds like the good days hehe... Would you mind helping me out with just one more problem? I have a test tomorrow and I'm trying to do every problem I can find in the book about probabilities....</p>
<p>so here goes:</p>
<p>One person in every 1000 of apopulation is known to have tuberculosis. When tested, there is a 95% chance that a person with tuberculosis will in fact register positive. What are the chances of a person from this population having tuberculosis will also be tested positive?</p>
<p>I thought it would be P(being tested positive given that you have tuberculosis)= p(positive and tube) /p(tube) = (1/1000* 0.95)/(1/1000)
but wouldn't this just be 95%...this just sounds too easy to be right , would you mind checking/correcting my process?</p>
<p>great ;),thanks Dk-blue-Falcon yay one more problem down, anyone else want to confirm that? just so I don't screw up a similar problem on tomorrows test.</p>
<p>The way the problem is worded, 95% looks right to me too.</p>
<p>If the question had read 'What are the chances that a person selected from this population will have tuberculosis and will also be tested positive?' , then the answer would be (1/1000) (0.95) .</p>
<p>Yea, since it used "also" in the question, I was having a hard time trying to understand what exactly it was looking for. Your revision makes it much more clear.</p>
<p>The 2nd equation optimizerdad posted is probably the right setup:
1 - (1-p)^3 = 1.2 (p^3)</p>
<p>The 1's cancel so you can factor out a p and solve a simple quadratic equation.</p>
<p>(actually the reason you can tell 1 -(1-p)^3 = 0.2 + p^3 is wrong is because it has imaginary roots--for example if you changed 0.2 to 0.8. But if you use (1+fail%)p^3 instead, your quadratic equation will always have a positive real solution.)</p>
<p>NicoPico:
Nice catch, I missed that one. As you pointed out, we would have a cubic equation, on the face of it, but it would swiftly reduce to a quadratic.</p>
<p>this question is too advanced for Statistics topic which is covered in the SAT. usually the statistics topic involves multiplying or adding probability. this questions seems inaccurate</p>
<p>yeah no i know, sorry. This isn't a real sat question, I just tend to post any mathematical doubt I have here so that all the 800M scorers and other mathematically apt people can help me out. sorry</p>
there is a 95% chance that a person with tuberculosis will register positive
is actually the answer to
What are the chances that a person from this population having tuberculosis will be tested positive?
Its 95%.</p>
<p>Optimizerdad optimized the question in a simple(x) ;) way:
</p>
<p>Lets try a similar question:
What are the chances that a randomly selected person will not have tuberculosis and will be tested positive?<a href="yikes!">/u</a>
Not enough information to answer it.</p>
<p>If we know that
When tested, there is a 1% chance that a person with no tuberculosis will register positive,
than we can answer that question:
1000 - 1 = 999
(999/1000) (0.01).</p>
<p>One more twist now. It makes a question a far cry from the SAT, but more realistic.</p>
<p>One person in every 1000 of a population is known to have tuberculosis. When tested, there is a 95% chance that a person with tuberculosis will in fact register positive (not a very reliable test).
Furthermore, there is a 1% chance that a person with no tuberculosis will also register positive (makes the test even sloppier).
If a person tested positive, what are the chances that this person has tuberculosis?</p>
<p>Let's make a test very precise and see what happens. One person in every 1000 of a population is known to have tuberculosis.
Test is 100% accurate for a person with tuberculosis.
There is a (1/9)% chance that a person with no tuberculosis will register positive.
If a person tested positive, what are the chances that s/he has tuberculosis?</p>
<p>Should be very high, right?
My answer - 47%.
Math defeats common sense.</p>