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Math Expert V
Joined: 02 Sep 2009
Posts: 60544
Andy and Bob play a game in which a computer randomly selects two real  [#permalink]

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12 00:00

Difficulty:   75% (hard)

Question Stats: 53% (02:30) correct 47% (03:19) wrong based on 45 sessions

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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

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VP  V
Joined: 19 Oct 2018
Posts: 1293
Location: India
Re: Andy and Bob play a game in which a computer randomly selects two real  [#permalink]

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6
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+1-a-b>0

(a-1)(b-1)>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%
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##### General Discussion
VP  P
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]

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1
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: 121-20=101
Probability: 101/121~83.4%

Ans (D)
Director  D
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]

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2
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(y-1) + (1-y) > 0
(y-1)(x-1) > 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

Senior Manager  P
Joined: 25 Jul 2018
Posts: 468
Re: Andy and Bob play a game in which a computer randomly selects two real  [#permalink]

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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 %

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GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: Andy and Bob play a game in which a computer randomly selects two real  [#permalink]

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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 Questions

I 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
36-8/36*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
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VP  V
Joined: 19 Oct 2018
Posts: 1293
Location: India
Re: Andy and Bob play a game in which a computer randomly selects two real  [#permalink]

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1
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 co-ordinate 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 Questions

I 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
36-8/36*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  P
Joined: 07 Mar 2019
Posts: 579
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: Andy and Bob play a game in which a computer randomly selects two real  [#permalink]

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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 co-ordinate 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 Questions

I 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
36-8/36*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

Kudos 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  P
Joined: 21 Jun 2017
Posts: 395
Location: India
Concentration: Finance, Economics
Schools: IIM
GMAT 1: 620 Q47 V30
GPA: 3
WE: Corporate Finance (Commercial Banking)
Re: Andy and Bob play a game in which a computer randomly selects two real  [#permalink]

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Between 0 - 10 does not mean 1-9 ????
Veritas Prep GMAT Instructor V
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]

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ShankSouljaBoi wrote:
Between 0 - 10 does not mean 1-9 ????

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