GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Feb 2019, 09:39

### 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 February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Algebra Webinar

February 17, 2019

February 17, 2019

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• ### Free GMAT Strategy Webinar

February 16, 2019

February 16, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# The probability of winning game A is X and the probability of winning

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52902
The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

29 Sep 2016, 03:38
1
1
6
00:00

Difficulty:

35% (medium)

Question Stats:

68% (01:29) correct 32% (01:16) wrong based on 259 sessions

### HideShow timer Statistics

The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?

A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y

_________________
SC Moderator
Joined: 13 Apr 2015
Posts: 1687
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
GPA: 4
WE: Analyst (Retail)
Re: The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

29 Sep 2016, 06:21
1
Probability of winning exactly once --> A wins and B loses or A loses and B wins
= X(1 - Y) + (1 - X)Y
= X - XY + Y - XY
= X + Y - 2XY

Current Student
Joined: 26 Jan 2016
Posts: 103
Location: United States
GPA: 3.37
Re: The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

29 Sep 2016, 11:54
I picked numbers

X=1/3
Y=1/4

Probability of winning A and losing B (1/3)*(3/4)=1/4
Probability of losing A and winning B (2/3)*(1/4)=1/6

Because it is either option A OR option B we add the two probabilities together

1/4+1/6=5/12. If you plug the numbers back into the answer choices you will get D.
Director
Status: Come! Fall in Love with Learning!
Joined: 05 Jan 2017
Posts: 542
Location: India
Re: The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

27 Feb 2017, 03:24
P(game A win) = X, P(game A lose) = 1-X
P(game B win) = Y, P(game B lose) = 1-Y
P(exactly one game win = P(game A win) P(game B lose) + P(game B win) P(game A lose) = X(1-Y) + Y(1-X) = X +Y -2XY. Option D
_________________

GMAT Mentors

Manager
Joined: 24 Dec 2016
Posts: 96
Location: India
Concentration: Finance, General Management
WE: Information Technology (Computer Software)
Re: The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

13 Mar 2017, 20:16
Bunuel wrote:
The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?

A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y

Hi,

I used the following approach :
P(Winning A) = X
P(Winning B) = Y
p(Winning Both) = X.Y

Total = X+Y-XY

Kindly suggest where I'm going wrong.
Intern
Joined: 10 Aug 2017
Posts: 2
Location: United States
Concentration: Finance, Technology
Re: The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

24 Jun 2018, 12:33
ByjusGMATapp wrote:
P(game A win) = X, P(game A lose) = 1-X
P(game B win) = Y, P(game B lose) = 1-Y
P(exactly one game win = P(game A win) P(game B lose) + P(game B win) P(game A lose) = X(1-Y) + Y(1-X) = X +Y -2XY. Option D

Confused why is this not X+y following gmatclub_dot_com/forum/the-probability-that-team-a-will-not-win-the-tournament-is-80-and-the-261771.html
Manager
Status: Student
Joined: 06 Jun 2018
Posts: 61
Location: India
Concentration: Finance, Marketing
GPA: 4
The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

24 Jun 2018, 12:55
Sguha0305 wrote:
ByjusGMATapp wrote:
P(game A win) = X, P(game A lose) = 1-X
P(game B win) = Y, P(game B lose) = 1-Y
P(exactly one game win = P(game A win) P(game B lose) + P(game B win) P(game A lose) = X(1-Y) + Y(1-X) = X +Y -2XY. Option D

Confused why is this not X+y following gmatclub_dot_com/forum/the-probability-that-team-a-will-not-win-the-tournament-is-80-and-the-261771.html

See in this question, two games are seperate. Each game have a seperate win and loss.
But in that link question its a tournament, there will be only a single win and multiple failures i.e. If team A is winning that automatically means that B is loosing

whereas in this question the win of team A doesnt effect team B. If team A wins, team B can also win.

Posted from my mobile device
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4915
Location: United States (CA)
Re: The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

12 Oct 2018, 06:59
Bunuel wrote:
The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?

A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y

P(winning exactly once) = 1 - P(winning zero times) - P(winning both times)

P(wWinning exactly once) = 1 - (1 - X)(1 - Y) - XY = 1 - (1 - X - Y + XY) - XY = X + Y - 2XY

Alternate Solution:

There are two ways of winning exactly once, either (Win A, Lose B) OR (Lose A, Win B). The probability of winning A is X, so the probability of losing A is (1 - X). Similarly, the probability of winning B is Y, and so the probability of losing B is (1 - Y).

P(Win A, Lose B) = X*(1 - Y) = X - XY

P(Lose A, Win B) = (1 - X)*Y = Y - XY

The probability that either of these events happens is the sum of their individual probabilities, so P(Winning exactly once) = X - XY + Y - XY = X + Y - 2XY

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 730
The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

12 Oct 2018, 12:31
Bunuel wrote:
The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?

A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y

$$? = P\left( {{\text{win}}\,\,{\text{just}}\,\,{\text{A}}\,\,{\text{or}}\,\,{\text{win}}\,\,{\text{just}}\,\,{\text{B}}} \right)\,\,\,\, = \,\,\,\,P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) - P\left( {{\text{win}}\,\,{\text{both}}} \right)$$

$$P\left( {{\rm{win}}\,\,{\rm{both}}} \right) = xy$$

$$\left[ {\,x,y\,\,{\rm{constants}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{win}}\,A\,\,{\rm{and}}\,\,{\rm{win}}\,B\,\,{\rm{are}}\,\,{\rm{independent}}\,\,{\rm{events}}\,} \right]$$

$$P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) = x + y - xy\,\,\,\,\,\left[ {\,{\text{simplifier}}\,} \right]$$

$$? = x + y - 2xy$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

"Kolmogorov - Poincaré - Gauss - Euler - Newton, are only five lives separating us from the source of our science."
_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13536
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: The probability of winning game A is X and the probability of winning  [#permalink]

### Show Tags

14 Oct 2018, 12:20
Hi All,

We're told that the probability of winning Game A is X and the probability of winning Game B is Y. We're asked when playing a SINGLE round of EACH of the two games, what is the probability of winning EXACTLY ONCE. This question can be solved in a couple of different ways, including by TESTing VALUES.

IF... X = .4 AND Y = .5
then there is a 40% chance to win Game A and a 60% chance to lose Game A
and a 50% chance to win Game B and a 50% chance to lose Game B.

The probability of winning A and losing B = (.4)(.5) = .2
The probability of winning B and losing A = (.5)(.6) = .3
Total probability of winning just one Game = .2 + .3 = .5

Thus, we're looking for an answer that equals .5 when X = .4 and Y = .5 There's only one answer that matches....

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Re: The probability of winning game A is X and the probability of winning   [#permalink] 14 Oct 2018, 12:20
Display posts from previous: Sort by