Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 60544

Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
04 Dec 2019, 00:03
Question Stats:
53% (02:30) correct 47% (03:19) wrong based on 45 sessions
HideShow timer Statistics
Competition Mode Question Andy and Bob play a game in which a computer randomly selects two real numbers between 0 and 10. Andy’s score is the sum of the numbers and Bob’s score is one more than the product of the numbers. If the person with the higher score wins the game, what is the probability that Bob wins? A. 18% B. 19% C. 68% D. 82% E. 90% Are You Up For the Challenge: 700 Level Questions
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




VP
Joined: 19 Oct 2018
Posts: 1293
Location: India

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
04 Dec 2019, 08:01
Andy and Bob play a game in which a computer randomly selects two real numbers between 0 and 10. Andy’s score is the sum of the numbers and Bob’s score is one more than the product of the numbers. If the person with the higher score wins the game, what is the probability that Bob wins? Let two numbers randomly selected by computer are a and b Bob will win if a*b+1>a+b ab+1ab>0 (a1)(b1)>0 either a>1 and b>1 or a<1 and b<1 Total possible area= 10*10= 100 Area which represents that Bob will win (Red highlighted portion)= 1*1+9*9= 82 Probability that Bob will win= 82/100= 0.82 or 82%
Attachments
Untitled.png [ 4.46 KiB  Viewed 848 times ]




VP
Joined: 24 Nov 2016
Posts: 1079
Location: United States

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
04 Dec 2019, 05:28
Quote: Andy and Bob play a game in which a computer randomly selects two real numbers between 0 and 10. Andy’s score is the sum of the numbers and Bob’s score is one more than the product of the numbers. If the person with the higher score wins the game, what is the probability that Bob wins?
A. 18% B. 19% C. 68% D. 82% E. 90% ASSUMING THAT REPETITION IS ALLOWED: ie. 00,11,22,33… {0,1…0,10}=10 ways AB+1≤A+B {1,1…1,10}=10 ways AB+1≤A+B Total cases with reps: 11*11=121 Winning cases with reps: 12120=101 Probability: 101/121~83.4% Ans (D)



Director
Joined: 30 Sep 2017
Posts: 576
GMAT 1: 720 Q49 V40
GPA: 3.8

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
04 Dec 2019, 08:54
Let's define 1st and 2nd numbers as x and y, where 0 < x,y < 10
Bob wins IF Bob > Andy, meaning that: xy+1 > x + y x(y1) + (1y) > 0 (y1)(x1) > 0
The above inequality (Bob>Andy) is satisfied ONLY IF: (1) 10>y>1 and 10>x>1 > Probability is: 9/10 * 9/10 = 81/100 (2) 0<y<1 and 0<x<1 > Probability is: 1/10 * 1/10 = 1/100 Thus, the probability that Bob wins is: (1) + (2) = 81/100 + 1/100 = 82/100 as the two conditions never overlap each other
Final answer is (D)



Senior Manager
Joined: 25 Jul 2018
Posts: 468

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
04 Dec 2019, 15:11
Let’s assume that two real numbers are a and b between 0 and 10.
—> Andy —(a+b) —> Bob —(ab+1 ) If the person with the higher score wins the game, what is the probability that Bob wins? —> a and b could be equal to each other —> Total combination 11* 11= 121
Bob wins: (ab +1) > a+ b a( b—1) > b—1 (a—1)(b—1) > 0
Case1: a >1 and b >1 —> 9*9 = 81 Case2: a< 1 and b<1 —> a and b could be equal to each other—> 1 (just one possibility—> a=0,b=0) Total possibilities in which Bob wins—> 81+1 = 82
The probability = 82/ 121= 0.677... —> 68 % The answer is C
Posted from my mobile device



Director
Joined: 07 Mar 2019
Posts: 579
Location: India
WE: Sales (Energy and Utilities)

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
05 Dec 2019, 02:54
Bunuel wrote: Competition Mode Question Andy and Bob play a game in which a computer randomly selects two real numbers between 0 and 10. Andy’s score is the sum of the numbers and Bob’s score is one more than the product of the numbers. If the person with the higher score wins the game, what is the probability that Bob wins? A. 18% B. 19% C. 68% D. 82% E. 90% Are You Up For the Challenge: 700 Level QuestionsI considered a and b as numbers from 1 to 9 since inclusive is not mentioned in the question. Also i took cases without repetition. However, I solved as follow: If for A number is 1, 2≤B≤9 i.e. 8 cases(without repetition) .... A .... 2, 3≤B≤9 i.e. 7 cases and so on... till 1 case for A as 8 So, total 36 cases are there in which 8 cases(where A is 1 and 2≤B≤9) are neutral i.e. both scores are equal. Since win percent for Bob is required, i took these cases as loss and calculated win percent as 368/36*100 = ~77% Here's my question: If cases where both A and B are equal(i.e. repetition) are considered B wins in 8 out of 9 cases taking win percentage to 36/45*100 = 80% Also if that's so, cases as discussed above are repeated let's say For B if number is 1 2≤A≤9 8 cases and SO ON... Though this does not make much difference but it creates lot more confusion. For that reason i estimated that my calculated should be lesser than ~77% and marked C. Can anyone help with this.
_________________
Ephemeral Epiphany..!
GMATPREP1 590(Q48,V23) March 6, 2019 GMATPREP2 610(Q44,V29) June 10, 2019 GMATPREPSoft1 680(Q48,V35) June 26, 2019



VP
Joined: 19 Oct 2018
Posts: 1293
Location: India

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
05 Dec 2019, 05:57
Read the question again. Computer can select any 2 real numbers, but you considered only integers; hence your solution is not correct. You gotta solve it with geometric probability. As we have 2 variables, find the total cases and favourable cases on coordinate plane. lnm87 wrote: Bunuel wrote: Competition Mode Question Andy and Bob play a game in which a computer randomly selects two real numbers between 0 and 10. Andy’s score is the sum of the numbers and Bob’s score is one more than the product of the numbers. If the person with the higher score wins the game, what is the probability that Bob wins? A. 18% B. 19% C. 68% D. 82% E. 90% Are You Up For the Challenge: 700 Level QuestionsI considered a and b as numbers from 1 to 9 since inclusive is not mentioned in the question. Also i took cases without repetition. However, I solved as follow: If for A number is 1, 2≤B≤9 i.e. 8 cases(without repetition) .... A .... 2, 3≤B≤9 i.e. 7 cases and so on... till 1 case for A as 8 So, total 36 cases are there in which 8 cases(where A is 1 and 2≤B≤9) are neutral i.e. both scores are equal. Since win percent for Bob is required, i took these cases as loss and calculated win percent as 368/36*100 = ~77% Here's my question: If cases where both A and B are equal(i.e. repetition) are considered B wins in 8 out of 9 cases taking win percentage to 36/45*100 = 80% Also if that's so, cases as discussed above are repeated let's say For B if number is 1 2≤A≤9 8 cases and SO ON... Though this does not make much difference but it creates lot more confusion. For that reason i estimated that my calculated should be lesser than ~77% and marked C. Can anyone help with this. Posted from my mobile device



Director
Joined: 07 Mar 2019
Posts: 579
Location: India
WE: Sales (Energy and Utilities)

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
05 Dec 2019, 06:34
nick1816 wrote: Read the question again. Computer can select any 2 real numbers, but you considered only integers; hence your solution is not correct. You gotta solve it with geometric probability. As we have 2 variables, find the total cases and favourable cases on coordinate plane. lnm87 wrote: Bunuel wrote: Competition Mode Question Andy and Bob play a game in which a computer randomly selects two real numbers between 0 and 10. Andy’s score is the sum of the numbers and Bob’s score is one more than the product of the numbers. If the person with the higher score wins the game, what is the probability that Bob wins? A. 18% B. 19% C. 68% D. 82% E. 90% Are You Up For the Challenge: 700 Level QuestionsI considered a and b as numbers from 1 to 9 since inclusive is not mentioned in the question. Also i took cases without repetition. However, I solved as follow: If for A number is 1, 2≤B≤9 i.e. 8 cases(without repetition) .... A .... 2, 3≤B≤9 i.e. 7 cases and so on... till 1 case for A as 8 So, total 36 cases are there in which 8 cases(where A is 1 and 2≤B≤9) are neutral i.e. both scores are equal. Since win percent for Bob is required, i took these cases as loss and calculated win percent as 368/36*100 = ~77% Here's my question: If cases where both A and B are equal(i.e. repetition) are considered B wins in 8 out of 9 cases taking win percentage to 36/45*100 = 80% Also if that's so, cases as discussed above are repeated let's say For B if number is 1 2≤A≤9 8 cases and SO ON... Though this does not make much difference but it creates lot more confusion. For that reason i estimated that my calculated should be lesser than ~77% and marked C. Can anyone help with this. Posted from my mobile deviceKudos to your solution. Realized it after i saw it why i got less than 82%
_________________
Ephemeral Epiphany..!
GMATPREP1 590(Q48,V23) March 6, 2019 GMATPREP2 610(Q44,V29) June 10, 2019 GMATPREPSoft1 680(Q48,V35) June 26, 2019



Senior Manager
Joined: 21 Jun 2017
Posts: 395
Location: India
Concentration: Finance, Economics
GPA: 3
WE: Corporate Finance (Commercial Banking)

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
16 Jan 2020, 04:55
Between 0  10 does not mean 19 ????



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9998
Location: Pune, India

Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
Show Tags
16 Jan 2020, 05:57
ShankSouljaBoi wrote: Between 0  10 does not mean 19 ???? With the use of the word "between", there is always a question of "inclusive" or "exclusive"  whether the limits are included or not. Here it doesn't matter because we are talking about real numbers. So 0.00000001 is certainly included and 0 is just one of infinite acceptable values. Assuming the numbers to be x and y, xy + 1 > x + y xy + 1  x  y > 0 x(y  1) 1(y  1) > 0 (x  1)(y  1) > 0 Either both factors should be positive (x > 1 and y > 1)  Probability = (9/10)*(9/10) = 81/10 or both factors should be negative (x < 1 and y < 1)  Probability = (1/10)*(1/10) = 1/100 Total probability = 81/100 + 1/100 = 82/100
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >




Re: Andy and Bob play a game in which a computer randomly selects two real
[#permalink]
16 Jan 2020, 05:57






