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Probability Question

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Probability Question

by gmatassistance » Sun Mar 28, 2010 4:57 pm
A machine consists of three components, which are components X, Y, and Z. These three components operate properly independently of one another. The machine operates properly only when component X operates properly and at least one of components Y and Z operates properly. The probability that component X operates properly is 0.8, the probability that component Y operates properly is 0.4, and the probability that component Z operates properly is 0.3. What is the probability that the machine operates properly?

0.096
0.252
0.464
0.583
0.648
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by gmatassistance » Sun Mar 28, 2010 4:59 pm
I tried using the probability formula where the Probability of A or B = pA x pB - (pA + pB) to figure out the probability for components Y or Z to operate properly however got a negative answer.

Can someone please explain why the above formula didn't work and what other strategies to keep in mind for such questions?

Thanks!
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by student22 » Sun Mar 28, 2010 7:57 pm
I'm not sure if this is the right way to solve this, but couldn't you take the probability of X occurring, 0.80 * (1 - PZ and PY NOT OCCURRING). 0.80 * (1 - (.7 * .6) = 0.80 * .58 = .464.

My reasoning is that its easier to find the the probability that they will both fail and then subtract that from 1. This gives the probability of 1 work or both working, which you multiply by X.

Hopefully this helps, if not, maybe someone has a better way.
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by pops » Sun Mar 28, 2010 9:53 pm
machine will work properly if
(A) works and (B or C) works
this means,
p = A On, B On, C On + A On, B On, C Off + A On, B Off, C On
= 0.8*0.4*0.3 + 0.8*0.4*(1-0.3) + 0.8*(1-0.4)*0.3
= 0.464
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by koolk86 » Sun Mar 28, 2010 10:13 pm
I have solved it like this :

For the machine to work properly X must work properly and ATLEAST Y or Z must work properly

Probability of X working properly = 0.8

Probability of Y working properly = 0.4

Probability of z working properly = 0.3

Probability of X not working properly = 0.2 (1- 0.8)

Probability of Y not working properly = 0.6 (1- 0.4)

Probability of Z not working properly = 0.7 (1- 0.3)


Now use the above:

P(X works) * P(Y works) * P (Z does not work) = 0.8 * 0.4 * 0.7 = 0.244

OR

P(X works)* P(Ydoes not work) * P(Z works) = 0.8 *0.6*0.3 = 0.144

OR

P(X works) * P ( Y works ) * P ( Z works) = 0.8 * 0.4 * 0.3 = 0.096

Add All you get : 0.464

This is a bit long method and involves calculations.

I guess one could also solve by 1-x method ???
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by hooliganpete » Mon Mar 29, 2010 12:56 pm
I found this method to be quick and painless:

Machine works when X works and either Y or Z so find the probability that Either Y or Z works. The 1- method is great for this first step.

Prob(X doesn't work) * Prob(Y doesn't work) = .42

1-.42 = .58

Thus, .58 = Prob(either X or Y works)

Now since X has to work, .42*.8 = Prob(X works)*Prob(Either X or Y work) = .464
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by eaakbari » Mon Mar 29, 2010 12:58 pm
Hey gmatassistance
I think the mistake you are making here is applying the probability law.
It states that all the events are independent.
The law which you are applying is only applicable for non - independent events, even mutually exclusive events would do but nope not independent.

For independent remember the multiplication rule which is

For A and B to both occur that equals P(A) *P(B)

For this question realise that there are 3 cases where machine will work
X and Y are operating but Z is not
X and Z are operating but Y is not
X,Y and Z are operating

So prob will be
0.8*0.4*0.7 + 0.8*0.3*0.6 + 0.8*0.3*0.6

Which equals 0.464

Hope this helps

E
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