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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, 04:16

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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

Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]

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30 Sep 2012, 04:27

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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 \(0.25+xy-0.25xy=0.55\) 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 \(y=0.8.\)
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]

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03 Oct 2012, 22:25

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 \(0.25+xy-0.25xy=0.55\) 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, 22:49

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 \(0.25+xy-0.25xy=0.55\) 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) ??

Re: Each of the variables X, Y, and Z can only be 0 or 1. The [#permalink]

Show Tags

03 Oct 2012, 23:36

1

This post received KUDOS

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 \(0.25+xy-0.25xy=0.55\) 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: \(P(A\cup{B})=P(A)+P(B)-P(A\cap{B}).\)
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.