Solve the problem.
-A fair spinner has regions numbered 1 through 21. What is the probability that the spinner will stop on an even number or a multiple of 3?
A) $\frac { 1 } { 3 }$
B) $\frac { 10 } { 9 }$
C) 17
D) $\frac { 2 } { 3 }$

Question

Solve the problem.
-A fair spinner has regions numbered 1 through 21. What is the probability that the spinner will stop on an even number or a multiple of 3? 
A) $\frac { 1 } { 3 }$
B) $\frac { 10 } { 9 }$
C) 17
D) $\frac { 2 } { 3 }$

Quizplus · Accepted Answer

The even numbers between 1 and 21 are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20, giving us 10 even numbers. The multiples of 3 between 1 and 21 are 3, 6, 9, 12, 15, 18, and 21, giving us 7 multiples of 3. However, 6, 12, and 18 are counted in both categories, so we don't double-count them. Adding the unique counts together, we get 10 (even numbers) + 4 (additional unique multiples of 3 not already counted as even) = 14. Therefore, the probability is 14 out of 21, which simplifies to $\frac{2}{3}$.

Solve the Problem 13\frac { 1 } { 3 }31​ B) 109\frac { 10 } { 9 }910​

Solve the Problem $\frac { 1 } { 3 }$ B) $\frac { 10 } { 9 }$