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

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

Reveal Answer

Answer

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 happens  with probability $\frac{1}{10}\cdot\frac{1}{10}=\frac{1}{100}$

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


One Response

  1. Anu says:

    I want to see the solution