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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
4
Kudos
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)
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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
1
Kudos
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

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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
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.
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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
1
Kudos
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
CEO
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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
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. :thumbup:
Realized it after i saw it why i got less than 82%
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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
Between 0 - 10 does not mean 1-9 ????
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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
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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
KarishmaB wrote:
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

­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?
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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
3
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Expert Reply
saurabhbajpai wrote:
KarishmaB wrote:
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

­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?

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.
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Re: Andy and Bob play a game in which a computer randomly selects two real [#permalink]
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