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

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)

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

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.

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

Kudos to your solution.
Realized it after i saw it why i got less than 82%

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.