Question

Two wheels are spun, and each wheel's pointer selects one number. The first wheel is divided into four equal regions numbered $3, 4, 5,$ and $8;$ the second wheel is divided into three equal regions numbered $6, 7,$ and $9.$ What is the probability that the sum of the two selected numbers is even?

$\dfrac{1}{6}$

$\dfrac{3}{7}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{5}{7}$

Solution:

The sum is even when both numbers are even or both are odd. The first wheel has evens $\{4, 8\}$ and odds $\{3, 5\},$ each with probability $\frac{2}{4} = \frac{1}{2}.$ The second wheel has even $\{6\}$ with probability $\frac{1}{3}$ and odds $\{7, 9\}$ with probability $\frac{2}{3}.$

So the probability of an even sum is $\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}.$

Thus, the correct answer is C .

1989 AMC 8 Problem 25

Solution:

Problem 25 in Other Years