let x and y numbers you will get when they start playing
Bob's equation is xy+1
Andy's equation is x+y
and 0<x<10,0<y<10

From question Bob wins if only his sum is greater than Andy's
=> xy+1>x+y
=> xy-x-y+1>0
=> (x-1)(y-1)>0
=>x-1>0,y-1>0 (or) x-1<0,y-1<0

consider x-1>0,y-1>0 & we know from question 0<x<10,0<y<10
so x,y have values from 2 to 10 to have above equations to be satisfied
=> we have 1<x<10, 1<y<10
9 intervals*9 intervals = 81 of total possible possibilities
(here i am not getting how to express this because x,y are real numbers)


consider x-1<0,y-1<0 & we know from question 0<x<10,0<y<10
so x,y have just one value 0<x<1,0<y<1 => we have 1 interval*1 interval= 1 of total possibilities

So on the whole we have 81+1 = 82 possibilities

And total number of possibilities are 10 intervals*10 intervals= 100

should be 82% IMO

this problem looks weirdly good one. OA please?

user123321

Total no of ways computer can select two nos = 11C2 = 55

Case 1: one of the numbers selected is 0
0,1 = No1 wins
0,2-10 = Andy wins

Case 2 : one fo the nos is 1
1,2 = No1 wins
1+n = 1*n + 1
No one wins

Case 3: one of the nos is 2
2,3
2+3 = 5
2*3 + 1 = 7
Bob wins
SAme for rest of the selections

No of selections in which Bob can win:
2,3-10 (8 pairs)
3,4-10(7pairs)
4,5-10(6pairs)
.
.
.

=> 8+7+..1 = 8(2+7)/2 = 36

Prob = 36/55 *100 = 65

Option C is the closest
So IMO:C

Crap.. i didnt notice between 0 and 10

user123321 wrote:let x and y numbers you will get when they start playing
Bob's equation is xy+1
Andy's equation is x+y
and 0<x<10,0<y<10

From question Bob wins if only his sum is greater than Andy's
=> xy+1>x+y
=> xy-x-y+1>0
=> (x-1)(y-1)>0
=>x-1>0,y-1>0 (or) x-1<0,y-1<0

consider x-1>0,y-1>0 & we know from question 0<x<10,0<y<10
so x,y have values from 2 to 10 to have above equations to be satisfied
=> we have 1<x<10, 1<y<10
9 intervals*9 intervals = 81 of total possible possibilities
(here i am not getting how to express this because x,y are real numbers)


consider x-1<0,y-1<0 & we know from question 0<x<10,0<y<10
so x,y have just one value 0<x<1,0<y<1 => we have 1 interval*1 interval= 1 of total possibilities

So on the whole we have 81+1 = 82 possibilities

And total number of possibilities are 10 intervals*10 intervals= 100

should be 82% IMO

this problem looks weirdly good one. OA please?

user123321

9*9 will get you total no of arrangements
while computer is just selecting two nos
3,5 is the same as 5,3

here i am not getting how to express that.

hmm...in other words i will try to explain
(x-1)(y-1)>0
x>1,y>1
so we have the product (x-1)(y-1) value ranging from 0 to 81 => 81 intervals

//ly x<1,y<1
so we have the product (x-1)(y-1) value ranging from 0 to 1=> 1 interval

but from given 0<x<10,0<y<10
so we have the product xy value ranging from 0 to 100 => 100 intervals

so probability = (81+1)/100 = 82%
but i am not so sure whether it is propoer or not.

user123321

karthikpandian19 wrote: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%

Let the two numbers = x and y.

P(Bob wins) = 1 - P(Adam wins).
(*Please see the disclaimer below.)

P(Adam wins):
Adam wins when x + y > xy + 1:
y-1 > xy - x
y-1 > x(y-1)
x(y-1) - (y-1) < 0
(y-1)(x-1) < 0.

The inequality is valid when y<1 and x>1 (yielding negative*positive) or y>1 and x<1 (yielding positive*negative).

Since the total range is from 0 to 10:
P(x<1 and y>1) = 1/10 * 9/10 = 9/100.
P(x>1 and y<1) = 9/10 * 1/10 = 9/100.
Since Adam wins in either case, we add the fractions:
P(Adam wins) = 9/100 + 9/100 = 18/100.

Thus, P(Bob wins) = 1 - 18/100 = 82/100.

The correct answer is D.

Disclaimer:
If either number = 1, the result will be a draw, since the sum will be equal to one more than the product:
x + 1 = x(1) + 1.
Since there are an infinite number of real numbers between 0 and 10, P(1) is infinitely small.
Thus, the solution above ignores this probability.

Nice explanation Mitch.

user123321