1991 AJHSME Problems/Problem 22


Each spinner is divided into $3$ equal parts. The results obtained from spinning the two spinners are multiplied. What is the probability that this product is an even number?

[asy] draw(circle((0,0),2)); draw(circle((5,0),2)); draw((0,0)--(sqrt(3),1)); draw((0,0)--(-sqrt(3),1)); draw((0,0)--(0,-2)); draw((5,0)--(5+sqrt(3),1)); draw((5,0)--(5-sqrt(3),1)); draw((5,0)--(5,-2)); fill((0,5/3)--(2/3,7/3)--(1/3,7/3)--(1/3,3)--(-1/3,3)--(-1/3,7/3)--(-2/3,7/3)--cycle,black); fill((5,5/3)--(17/3,7/3)--(16/3,7/3)--(16/3,3)--(14/3,3)--(14/3,7/3)--(13/3,7/3)--cycle,black); label("$1$",(0,1/2),N); label("$2$",(sqrt(3)/4,-1/4),ESE); label("$3$",(-sqrt(3)/4,-1/4),WSW); label("$4$",(5,1/2),N); label("$5$",(5+sqrt(3)/4,-1/4),ESE); label("$6$",(5-sqrt(3)/4,-1/4),WSW); [/asy]

$\text{(A)}\ \frac{1}{3} \qquad \text{(B)}\ \frac{1}{2} \qquad \text{(C)}\ \frac{2}{3} \qquad \text{(D)}\ \frac{7}{9} \qquad \text{(E)}\ 1$


Instead of computing this probability directly, we can find the probability that the product is odd, and subtract that from $1$.

The product of two integers is odd if and only if each of the two integers is odd. The probability the first spinner yields an odd number is $2/3$ and the probability the second spinner yields an odd number is $1/3$, so the probability both yield an odd number is $(2/3)(1/3)=2/9$.

The desired probability is thus $1-2/9=7/9\rightarrow \boxed{\text{D}}$.

See Also

1991 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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