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Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]
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30 Sep 2012, 05:16
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Each of the variables X, Y, and Z can only be 0 or 1. The values of the three variables are independent of one another. The probability that Z is 1 is .25. The probability that X is 1 is less than the probability that Y is 1. The probability that XY + Z is at least 1 is .55.
The probability that X is 1 The probability that Y is 1 .35 .50 .60 .75 .80 .90
Answer : The probability that X is 1 : .50 The probability that Y is 1 : .80



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Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]
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30 Sep 2012, 05:27
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smartass666 wrote: Each of the variables X, Y, and Z can only be 0 or 1. The values of the three variables are independent of one another. The probability that Z is 1 is .25. The probability that X is 1 is less than the probability that Y is 1. The probability that XY + Z is at least 1 is .55.
The probability that X is 1 The probability that Y is 1 .35 .50 .60 .75 .80 .90
Answer : The probability that X is 1 : .50 The probability that Y is 1 : .80 \(P(XY+Z \geq1) = P(XY=1)+P(Z=1)P(XYZ=1).\) If \(P(X=1)=x, \,P(Y=1)=y,\) then \(0.25+xy0.25xy=0.55\) from which \(xy=0.4.\) If this is one of those table questions, where we have to choose one value for each from the given list, knowing that \(x<y\), we have \(x=0.5\) and \(y=0.8.\)
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Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]
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03 Oct 2012, 23:25
EvaJager wrote: smartass666 wrote: Each of the variables X, Y, and Z can only be 0 or 1. The values of the three variables are independent of one another. The probability that Z is 1 is .25. The probability that X is 1 is less than the probability that Y is 1. The probability that XY + Z is at least 1 is .55.
The probability that X is 1 The probability that Y is 1 .35 .50 .60 .75 .80 .90
Answer : The probability that X is 1 : .50 The probability that Y is 1 : .80 \(P(XY+Z \geq1) = P(XY=1)+P(Z=1)P(XYZ=1).\) If \(P(X=1)=x, \,P(Y=1)=y,\) then \(0.25+xy0.25xy=0.55\) from which \(xy=0.4.\) If this is one of those table questions, where we have to choose one value for each from the given list, knowing that \(x<y\), we have \(x=0.5\) and \(y=0.8.\) Can you please explain how you arrived at that equation?? Can it apply to an inequality ie (xy + z >= 1) ??
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Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]
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03 Oct 2012, 23:49
MacFauz wrote: EvaJager wrote: smartass666 wrote: Each of the variables X, Y, and Z can only be 0 or 1. The values of the three variables are independent of one another. The probability that Z is 1 is .25. The probability that X is 1 is less than the probability that Y is 1. The probability that XY + Z is at least 1 is .55.
The probability that X is 1 The probability that Y is 1 .35 .50 .60 .75 .80 .90
Answer : The probability that X is 1 : .50 The probability that Y is 1 : .80 \(P(XY+Z \geq1) = P(XY=1)+P(Z=1)P(XYZ=1).\) If \(P(X=1)=x, \,P(Y=1)=y,\) then \(0.25+xy0.25xy=0.55\) from which \(xy=0.4.\) If this is one of those table questions, where we have to choose one value for each from the given list, knowing that \(x<y\), we have \(x=0.5\) and \(y=0.8.\) Can you please explain how you arrived at that equation?? Can it apply to an inequality ie (xy + z >= 1) ?? It is provided in the question stem...



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Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]
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04 Oct 2012, 00:36
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MacFauz wrote: EvaJager wrote: smartass666 wrote: Each of the variables X, Y, and Z can only be 0 or 1. The values of the three variables are independent of one another. The probability that Z is 1 is .25. The probability that X is 1 is less than the probability that Y is 1. The probability that XY + Z is at least 1 is .55.
The probability that X is 1 The probability that Y is 1 .35 .50 .60 .75 .80 .90
Answer : The probability that X is 1 : .50 The probability that Y is 1 : .80 \(P(XY+Z \geq1) = P(XY=1)+P(Z=1)P(XYZ=1).\) If \(P(X=1)=x, \,P(Y=1)=y,\) then \(0.25+xy0.25xy=0.55\) from which \(xy=0.4.\) If this is one of those table questions, where we have to choose one value for each from the given list, knowing that \(x<y\), we have \(x=0.5\) and \(y=0.8.\) Can you please explain how you arrived at that equation?? Can it apply to an inequality ie (xy + z >= 1) ?? \(X,Y,Z\) can be either \(0\) or \(1\). Possible values of the expression \(XY+Z\) are \(0, 1\) and \(2.\) The greatest value of \(XY+Z\) is \(2\), for \(X=Y=Z=1\), and it is equal to \(1\) when either \(XY=1\) or \(Z=1.\) We have to compute the probability \(P(XY=1 \,\,or \,Z=1)\) for which we use the formula of the union: \(P(A\cup{B})=P(A)+P(B)P(A\cap{B}).\)
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Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]
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11 Dec 2012, 08:43
Hi here's my attempt for the question: "probability that XY + Z is at least 1 is .55" means that either XY or Z is 1 or both are 1
P(XY=0,Z=1)+P(XY=1,Z=0)+P(XY=1,Z=1)=0.55 P(XY=0,Z=0) = 10.55=0.45 =>(both XY and Z are 0)
we can find out P(XY=0)=0.45/0.75=0.6
so, 0.6 is the probability that XY are 0
The combination of X,Y are as follows:
X Y 0 0 0 1 1 0 1 1
For the first 3 cases the result of XY is 0, so P(X=1,Y=1) is 0.4
Just find from the table the combination that will fit into the equation P(X=1)P(Y=1)=0.4, and X<Y



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Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]
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19 Jan 2013, 00:08
difficult one...can this type of ques come in real gmat?
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