# Problem of the Day

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 Â higher score wins the game, what is the probability that Bob wins?

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

D. 82%

See the Solution

### Solution

[latexpage]

Two numbers are selected randomly from the interval $(0,10)$. Let the first number selected by $x$ and the second number be $y$. Andy’s score will be $x+y$ and Bob’s score will be $xy+1$. Bob will be the winner when $xy+1>x+y$. Solving this inequality:

$xy+1>x+y$

$xy-x-y+1>0$

$x(y-1)-(y-1)>0$

$(x-1)(y-1)>0$

This will be true if either $(x-1)$ and $(y-1)$ are both positive OR if $(x-1)$ and $(y-1)$ are both negative:

$x-1>0$ AND $y-1>0$

$x>1$ AND $y>1$

Because the numbers chosen are between 0 and 10, each number has a $\frac{9}{10}$ probability of being greater than one. The probability that they both are greater than one is $\frac{9}{10}\cdot\frac{9}{10}=\frac{81}{100}$.

The other possibility is that both numbers are less than 1, which, through similar analysis we can show happensÂ Â with probabilityÂ $\frac{1}{10}\cdot\frac{1}{10}=\frac{1}{100}$

Adding the two probabilities gives us $\frac{82}{100}$ or 82%

## One Response

1. Anu says:

I want to see the solution