Each of the variables X, Y, and Z can only be 0 or 1. The : Integrated Reasoning (IR)
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Each of the variables X, Y, and Z can only be 0 or 1. The

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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
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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 $$y=0.8.$$
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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, 22:25
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) ??

It is provided in the question stem...
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03 Oct 2012, 23:36
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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 $$0.25+xy-0.25xy=0.55$$ 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: $$P(A\cup{B})=P(A)+P(B)-P(A\cap{B}).$$
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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, 07:43
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) = 1-0.55=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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18 Jan 2013, 23:08
difficult one...can this type of ques come in real gmat?
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Re: Each of the variables X, Y, and Z can only be 0 or 1. The   [#permalink] 18 Jan 2013, 23:08
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