Andy and Bob play a game in which a computer randomly selects two real

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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%

A. 18%
B. 19%
C. 68%
D. 82%
E. 90%

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nick1816 · Accepted Answer

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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KarishmaB · Answer

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

exc4libur · Answer

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)

freedom128 · Answer

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

Final answer is (D)

lacktutor · Answer

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

unraveled · Answer

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.

nick1816 · Answer

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.

Posted from my mobile device

unraveled · Answer

Kudos to your solution.

Realized it after i saw it why i got less than 82%

ShankSouljaBoi · Answer

Between 0 - 10 does not mean 1-9 ????

saurabhbajpai · Answer

If the range includes both 0 and 10, then shouldn't the total number in the denominator be 11? How did we land at 10?

Bunuel · Answer

It does not matter whether 0 and 10 are included or not; that's not the point.

If x ranges from 0 to 10, the probability that it falls between 1 and 10 is 9/10. This is because the segment from 1 to 10, which measures 9 units, is 9/10 of the total range from 0 to 10, which is 10 units.

soma7 · Answer

Hi guys,

It might be difficult to think with variables at the instant in the exam. So, we will go with the conventional method. Translating each word in the question.

A=a+b
B=ab+1

P(bob winning) = 1 - p(Bob not winning)
Bob will not win only if one of the numbers ( a or b ) is 0 and the other being 2 to 9.
Total favourable outcomes for this = 1*8C1 = 8.
Total outcomes = 10C2 = 45.
therefore, 1 - 8/45 = 37/45 = approx. 82%

CrackGmat25 · Answer

Did it in 1 min 14 sec,

I never go with long calculations like others have done. After reading the question I understand the its weight and its extra hard for GMAT. Even if this appears in the actual exam I will prefer option elemination rather than finding ways to solve.

You can easily tell multiplication of two numbers would mostly give higher result than addition but for how many cases?

More than 50 ofcourse.  A & B eliminated.

Will it be slightly more than 50% like 68%? No, it will defninetly be higher than that. 5+6 and 5*6 numbers above than that will definietly follow.  C eliminated .

Now the choise is between D and E.

If you have done this far you are now at 50% chance of hitting right answer which was 20% at the start and thats a progress and you can do this in first 40 sec.

Now think, 90% is a very easy number it means in 0-10 i.e. 121 real number pairs there will be only 12 which will have sum higher than product seems bit less doesn't it? See 9+9 will be higher than 0*(any number) + 1. So 12 is bit less.  E eliminated

D is the best choise.

If the question weight this heavy don't solve play smart you will still be correct in 90% of the cases.

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