Each of the variables X, Y, and Z can only be 0 or 1. The values of

Question

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

EvaJager · Accepted Answer

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.

MacFauz · Answer

Can you please explain how you arrived at that equation?? Can it apply to an inequality ie (xy + z >= 1) ??

smartass666 · Answer

It is provided in the question stem...

EvaJager · Answer

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}).

sosohaihai · Answer

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

crimson_noise · Answer

Shouldn’t it be
P(XY+Z >= 1) = P(XY=1) + P(Z=1) + P(XYZ=1)

Bunuel · Answer

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).

The probability that X is 1	The probability that Y is 1	Values
		0.35
		0.50
		0.60
		0.75
		0.80
		0.90

Each of the variables X, Y, and Z can only be 0 or 1. The values of

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GMAT Two-part Analysis (TPA) Questions

Each of the variables X, Y, and Z can only be 0 or 1. The values of

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