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805+ Level|   Math Related|      
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smartass666
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Each of the variables X, Y, and Z can only be 0 or 1. The values of Updated on: 03 Oct 2023, 02:29
Originally posted by smartass666 on 30 Sep 2012, 05:16.
Last edited by BottomJee on 03 Oct 2023, 02:29, edited 3 times in total.
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The probability
that X is 1: 0.50 The probability
that Y is 1: 0.80
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Question Stats:

26% (02:58) correct 74% (02:51) wrong based on 219 sessions

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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 		Values
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0.50
0.60
0.75
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0.90
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The probability
that X is 1: 0.50 The probability
that Y is 1: 0.80
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EvaJager
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Each of the variables X, Y, and Z can only be 0 or 1. The values of 30 Sep 2012, 05:27
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smartass666
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
Show more

\(P(XY+Z \geq1) = P(XY=1)+P(Z=1)-P(XYZ=1).\)
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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Each of the variables X, Y, and Z can only be 0 or 1. The values of 03 Oct 2012, 23:25
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EvaJager
smartass666
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 \(y=0.8.\)
Show more

Can you please explain how you arrived at that equation?? Can it apply to an inequality ie (xy + z >= 1) ??
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Each of the variables X, Y, and Z can only be 0 or 1. The values of 03 Oct 2012, 23:49
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EvaJager
smartass666
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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It is provided in the question stem...
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Each of the variables X, Y, and Z can only be 0 or 1. The values of 04 Oct 2012, 00:36
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MacFauz
EvaJager
smartass666
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) ??
Show more

\(X,Y,Z\) 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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Each of the variables X, Y, and Z can only be 0 or 1. The values of 11 Dec 2012, 08:43
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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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Each of the variables X, Y, and Z can only be 0 or 1. The values of 14 Oct 2025, 10:15
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Shouldn’t it be
P(XY+Z >= 1) = P(XY=1) + P(Z=1) + P(XYZ=1)
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Each of the variables X, Y, and Z can only be 0 or 1. The values of 15 Oct 2025, 00:55
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crimson_noise
Shouldn’t it be
P(XY+Z >= 1) = P(XY=1) + P(Z=1) + P(XYZ=1)
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No. That formula double-counts.

XY + Z ≥ 1 means XY = 1 (and Z can be any value) or Z = 1 (and XY can be any value).

So the correct expression is:
P(XY + Z ≥ 1) = P(XY = 1) + P(Z = 1) - P(XY = 1 and Z = 1).

Since X, Y, and Z are independent,
P(XY = 1 and Z = 1) = P(XY = 1) * P(Z = 1).

Substitute this:
P(XY + Z ≥ 1) = P(XY = 1) + P(Z = 1) - P(XY = 1) * P(Z = 1).

Now rearrange:
= P(Z = 1) + P(XY = 1)(1 - P(Z = 1)).

This shows that the probability equals the probability that Z = 1 plus, when Z = 0, the probability that XY = 1. That’s why the correct form is:

P(Z = 1) + (1 - P(Z = 1)) * P(XY = 1),

not the one with + P(XYZ = 1).
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