Last visit was: 27 Sep 2026, 14:01 It is currently 27 Sep 2026, 14:01
Home
Messages
Tests
Schools
Events
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
805+ (Hard)|   Math Related|      
User avatar
smartass666
User avatar
Joined: 11 Feb 2012
Last visit: 02 Apr 2013
Posts: 10
Own Kudos:
512
 [35]
Given Kudos: 11
Posts: 10
Kudos: 512
 [35]
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.
Moved to new forum
Kudos
Add Kudos
35
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide Answer
Hide
Show
History
The probability
that X is 1: 0.50 The probability
that Y is 1: 0.80
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

27% (02:51) correct 73% (02:49) wrong based on 259 sessions

History

Date
Time
Result
Not Attempted Yet
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
0.35
0.50
0.60
0.75
0.80
0.90
Submit Answer
Start the Timer above, select the radio buttons, and click "Submit" to add this question to your Error log.
ShowHide Answer
Official Answer
The probability
that X is 1: 0.50 The probability
that Y is 1: 0.80
Most Helpful Reply
User avatar
EvaJager
User avatar
Joined: 22 Mar 2011
Last visit: 31 Aug 2016
Posts: 512
Own Kudos:
2,406
 [13]
Given Kudos: 43
WE:Science (Education)
Posts: 512
Kudos: 2,406
 [13]
Each of the variables X, Y, and Z can only be 0 or 1. The values of 30 Sep 2012, 05:27
7
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
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 \(y=0.8.\)
General Discussion
User avatar
MacFauz
User avatar
Joined: 02 Jul 2012
Last visit: 19 Mar 2022
Posts: 988
Own Kudos:
3,454
Given Kudos: 116
Location: India
Concentration: Strategy
Schools: Ross '16 (D) McCombs '16 (D) Kenan-Flagler '16 (WA) Olin '16 (M$) Tepper '16 (D)
GMAT 1: 740 Q49 V42
GPA: 3.8
WE:Engineering (Energy)
Schools: Ross '16 (D) McCombs '16 (D) Kenan-Flagler '16 (WA) Olin '16 (M$) Tepper '16 (D)
GMAT 1: 740 Q49 V42
Posts: 988
Kudos: 3,454
Each of the variables X, Y, and Z can only be 0 or 1. The values of 03 Oct 2012, 23:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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) ??
User avatar
smartass666
User avatar
Joined: 11 Feb 2012
Last visit: 02 Apr 2013
Posts: 10
Own Kudos:
512
 [1]
Given Kudos: 11
Posts: 10
Kudos: 512
 [1]
Each of the variables X, Y, and Z can only be 0 or 1. The values of 03 Oct 2012, 23:49
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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

It is provided in the question stem...
User avatar
EvaJager
User avatar
Joined: 22 Mar 2011
Last visit: 31 Aug 2016
Posts: 512
Own Kudos:
2,406
 [3]
Given Kudos: 43
WE:Science (Education)
Posts: 512
Kudos: 2,406
 [3]
Each of the variables X, Y, and Z can only be 0 or 1. The values of 04 Oct 2012, 00:36
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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: \(P(A\cup{B})=P(A)+P(B)-P(A\cap{B}).\)
avatar
sosohaihai
avatar
Joined: 28 Nov 2012
Last visit: 21 Aug 2013
Posts: 1
Own Kudos:
2
 [2]
Posts: 1
Kudos: 2
 [2]
Each of the variables X, Y, and Z can only be 0 or 1. The values of 11 Dec 2012, 08:43
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
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
User avatar
crimson_noise
User avatar
Joined: 14 May 2024
Last visit: 14 May 2026
Posts: 58
Own Kudos:
24
 [1]
Given Kudos: 14
GMAT Focus 1: 615 Q80 V85 DI77
GMAT Focus 1: 615 Q80 V85 DI77
Posts: 58
Kudos: 24
 [1]
Each of the variables X, Y, and Z can only be 0 or 1. The values of 14 Oct 2025, 10:15
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shouldn’t it be
P(XY+Z >= 1) = P(XY=1) + P(Z=1) + P(XYZ=1)
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 Sep 2026
Posts: 113,953
Own Kudos:
844,032
Given Kudos: 111,851
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,953
Kudos: 844,032
Each of the variables X, Y, and Z can only be 0 or 1. The values of 15 Oct 2025, 00:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
crimson_noise
Shouldn’t it be
P(XY+Z >= 1) = P(XY=1) + P(Z=1) + P(XYZ=1)
Show more
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).
Moderators:
Bunuel
Math Expert
113953 posts
FrozenDragon
DI Forum Moderator
409 posts