Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 28 Jul 2016, 05:57

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Intern
Joined: 11 Feb 2012
Posts: 12
Followers: 0

Kudos [?]: 42 [0], given: 11

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

### Show Tags

30 Sep 2012, 05:16
1
This post was
BOOKMARKED
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
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 90

Kudos [?]: 806 [1] , given: 43

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

### Show Tags

30 Sep 2012, 05:27
1
KUDOS
2
This post was
BOOKMARKED
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.

Moderator
Joined: 02 Jul 2012
Posts: 1230
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Followers: 107

Kudos [?]: 1223 [0], given: 116

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

### Show Tags

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 $$0.25+xy-0.25xy=0.55$$ 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) ??
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Intern
Joined: 11 Feb 2012
Posts: 12
Followers: 0

Kudos [?]: 42 [0], given: 11

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

### Show Tags

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 $$0.25+xy-0.25xy=0.55$$ 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...
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 90

Kudos [?]: 806 [1] , given: 43

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

### Show Tags

04 Oct 2012, 00:36
1
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.

Intern
Joined: 28 Nov 2012
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

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

### Show Tags

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) = 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
Intern
Joined: 19 May 2012
Posts: 36
Location: India
GMAT Date: 03-03-2014
WE: Information Technology (Computer Software)
Followers: 1

Kudos [?]: 7 [0], given: 0

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

### Show Tags

19 Jan 2013, 00:08
difficult one...can this type of ques come in real gmat?
_________________

Thanks
crazy4priya
GMATPrep 1 710/Q49/V38
GMATPrep 2 690/Q49/V34
Veritas Prep 700/Q50/V36/IR5
MGMT Test 1 700/Q51/V35/IR3

Re: Each of the variables X, Y, and Z can only be 0 or 1. The   [#permalink] 19 Jan 2013, 00:08
Similar topics Replies Last post
Similar
Topics:
1 x and y are expressed as decimals. 6 10 Jul 2014, 09:32
For integers, x and y, x=y+2. From the table below 1 06 Jul 2014, 05:45
2 The function f(x,y) is such that f(x,y) = (3x + 7y)/(x 4 07 May 2013, 02:14
1 IR question from Princeton Review 2013.. can some1 help 4 07 Jun 2012, 22:52
this is the question on gmat official site.. can some1 help 8 06 Jun 2012, 12:23
Display posts from previous: Sort by