Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 19 Oct 2013
Posts: 9
Location: United States
Concentration: Finance, Technology
GMAT Date: 11062013
GPA: 3.5
WE: Engineering (Investment Banking)

Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
31 Oct 2013, 11:18
Question Stats:
71% (02:34) correct 29% (02:36) wrong based on 275 sessions
HideShow timer Statistics
Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose? A. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 49436

Re: Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
31 Oct 2013, 11:25
Puneethrao wrote: Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
A. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 P = P(A wins, B wins, C loses) + P(A wins, B loses, C wins) + P(A loses, B wins, C wins) = 1/5*3/8*5/7 + 1/5*5/8*2/7 + 4/5*3/8*2/7 = 7/40. Answer: E.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Intern
Joined: 14 Aug 2013
Posts: 34
Location: United States
Concentration: Finance, Strategy
GMAT Date: 10312013
GPA: 3.2
WE: Consulting (Consumer Electronics)

Re: Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
31 Oct 2013, 11:32
Puneethrao wrote: Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
A. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 probability that exactly two of the three players will win = probability of (A will win,B will win,C will lose+ A will win , B will lose ,C will win+ A will lose,B will win,C will win) =>(1/5 * 3/8 * (1 2/7)) + (1/5 * (1  3/8) * 2/7) + ((1  1/5) * 3/8 * 2/7) =>15/280 + 10/280 + 24/280 => 49/280 =>7/40



Senior Manager
Status: Math is psychological
Joined: 07 Apr 2014
Posts: 421
Location: Netherlands
GMAT Date: 02112015
WE: Psychology and Counseling (Other)

Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
05 Mar 2015, 11:30
Well. there are three possible outcomes of interest, if 2 of the three have to win and one to lose:
AB...C AC...B CB...A
For the wins, we use the given probabilities. For the loss we use the remaining of the given probability. We multiply these together:
1/5 * 3/8 * 5/7 = 15 / 280 1/5 * 2/7 * 5/8 = 10 / 280 3/8 * 2/7 * 4/5 = 24 / 280
We now want to add these individual probabilities, which gives us: 49 / 280 = 7 / 40 ANS E



Director
Joined: 12 Nov 2016
Posts: 758
Location: United States
GPA: 2.66

Re: Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
16 Apr 2017, 20:35
Puneethrao wrote: Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
A. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 In order to solve this question we should set up the equation Prob (Exactly #success Exactly # failures)= P(A success x B success x C failure) + P(A success B failure C success) + P(A failure x B success x C success) Prob(Exactly # 2 success and Exactly 1 Failure)= P( 1/5 x 3/8 x 5/7) + P(1/5 x 5/8 x 2/7) + P( 4/5 x 3/8 x 2/7) = 49/280 Thus 7/40



eGMAT Representative
Joined: 04 Jan 2015
Posts: 2014

Re: Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
16 Apr 2017, 21:16
Puneethrao wrote: Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
A. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 To solve this question we need to consider three different cases  • A and B wins and C loses • The probability of the above case can be written as 
o \(P(A_wB_wC_L) = \frac{1}{5} * \frac{3}{8} * (1\frac{2}{7}) = \frac{3}{56}\) • B and C wins and A loses • The probability of the above case can be written as 
o \(P(A_LB_wC_w) = (1\frac{1}{5}) * \frac{3}{8} * \frac{2}{7} = \frac{3}{35}\) • C and A wins and B loses • The probability of the above case can be written as 
o \(P(A_wB_LC_w) = \frac{1}{5} * (1\frac{3}{8}) * \frac{2}{7} = \frac{1}{28}\) • The overall probability \(= P(A_wB_wC_L) + P(A_LB_wC_w) + P(A_wB_LC_w)\)
\(= \frac{3}{56} + \frac{3}{35} + \frac{1}{28}\) \(= \frac{3}{56} + \frac{3}{35} + \frac{1}{28}\) \(= \frac{(15+24+10)}{280}\) \(=\frac{7}{40}\) Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
_________________
Register for free sessions Number Properties  Algebra Quant Workshop
Success Stories Guillermo's Success Story  Carrie's Success Story
Ace GMAT quant Articles and Question to reach Q51  Question of the week
Must Read Articles Number Properties – Even Odd  LCM GCD  Statistics1  Statistics2 Word Problems – Percentage 1  Percentage 2  Time and Work 1  Time and Work 2  Time, Speed and Distance 1  Time, Speed and Distance 2 Advanced Topics Permutation and Combination 1  Permutation and Combination 2  Permutation and Combination 3  Probability Geometry Triangles 1  Triangles 2  Triangles 3  Common Mistakes in Geometry Algebra Wavy line  Inequalities Practice Questions Number Properties 1  Number Properties 2  Algebra 1  Geometry  Prime Numbers  Absolute value equations  Sets
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12450
Location: United States (CA)

Re: Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
05 Feb 2018, 12:33
Hi All, Since we have the probabilities of each person winning a game, we can 'map out' the 3 situations in which 2 of them win and 1 of them loses: A wins, B wins, C loses = (1/5)(3/8)(5/7) = 15/240 A wins, B loses, C wins = (1/5)(5/8)(2/7) = 10/240 A loses, B wins, C wins = (4/5)(3/8)(2/7) = 24/240 Total probability = 15/280 + 10/280 + 24/280 = 49/280 = 7/40 Final Answer: 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
CoFounder & 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!***********************



Manager
Joined: 02 Jul 2016
Posts: 116

Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
01 Jul 2018, 07:29
Bunuel wrote: Puneethrao wrote: Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
A. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 P = P(A wins, B wins, C loses) + P(A wins, B loses, C wins) + P(A loses, B wins, C wins) = 1/5*3/8*5/7 + 1/5*5/8*2/7 + 4/5*3/8*2/7 = 7/40. Answer: E. Hi Bunuel, Thanks for the explanation. I am repeatedly getting confused when to arrange data and when not to while finding out the probability. Here we just found out the probability of winning and loosing and multiplied it because they were independent events but we didn't arrange them because order of winning does not matter it. But in some of the questions, say the one below, we are also arranging the probability for one event. If a fair 6sided die is rolled three times, what is the probability that exactly one 3 is rolled?
25/216 50/216 25/72 25/36 5/6So how to determine whether the arrangement is required or not???? Thanks



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3528
Location: United States (CA)

Re: Alice, Benjamin, and Carol each try independentl
[#permalink]
Show Tags
04 Jul 2018, 19:07
Puneethrao wrote: Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
A. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 We must individually consider each possible outcome of having two winners and one loser. If Alice and Benjamin win and Carol loses we have: 1/5 x 3/8 x 5/7 = 1 x 3/8 x 1/7 = 3/56 If Alice and Carol win and Benjamin loses we have: 1/5 x 5/8 x 2/7 = 1 x 1/8 x 2/7 = 2/56 If Benjamin and Carol win and Alice loses we have: 4/5 x 3/8 x 2/7 = 1/5 x 3/2 x 2/7 = 1/5 x 3 x 1/7 = 3/35 Therefore, the probability that two of them will win and one will lose is: 3/56 + 2/56 + 3/35 = 15/280 + 10/280 + 24/280 = 49/280 = 7/40 Answer: E
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: Alice, Benjamin, and Carol each try independentl &nbs
[#permalink]
04 Jul 2018, 19:07






