2011 AMC 10B Problems/Problem 13


Two real numbers are selected independently at random from the interval $[-20, 10]$. What is the probability that the product of those numbers is greater than zero?

$\textbf{(A)}\ \frac{1}{9} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{4}{9} \qquad\textbf{(D)}\ \frac{5}{9} \qquad\textbf{(E)}\ \frac{2}{3}$


For the product of two numbers to be greater than zero, they either have to both be negative or both be positive. The interval for a positive number is $\frac{1}{3}$ of the total interval, and the interval for a negative number is $\frac{2}{3}$. Therefore, the probability the product is greater than zero is \[\frac{1}{3} \times \frac{1}{3} + \frac{2}{3} \times \frac{2}{3} = \frac{1}{9} + \frac{4}{9} = \boxed{\textbf{(D)} \frac{5}{9}}\]

Solution 2

We can also plot this problem on the coordinate grid. First, plot the possible region [the square with endpoints $(10,10), (-20, 10), (-20, -20), (-20, 10)$] Then, we can find our possible region which is if both numbers are positive or if both numbers are negative. Lets take the first case in which both numbers are positive: We can plot this as the top right of our possible square [$(10,10) ,(0,0) ,(10,0) ,(0,10)$]. Similarly if both numbers are negative we get the square [$(-20,-20), (0,0), (-20,0) ,(0,-20)$] Then we find the area of the possible region $30\cdot 30 = 900$ and then find the areas of our positive regions $10\cdot 10=100$ and $20\cdot 20 = 400$. Adding and simplifying we get $\frac{100+400}{900}=\frac{5}{9}$.

Solution 3

We will use complementary counting. The probability that the product is negative can be found by finding the probability that one number is positive and the other number is negative. The probability of a positive number being selected is $\frac13$, and the probability of a negative number being selected is $\frac23$. So, we see that the probability of the product being negative is \[2 \cdot \frac23 \cdot \frac13 = \frac49.\]

We multiply by $2$ because you can either pick the negative or positive number first.

Thus, the probability of the product being positive is \[1-\frac49 = \boxed{\textbf{(D) }\frac59}\]

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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