Question

a spinner has equal regions numbered 1 through 21. what is the probability that the spinner will stop on an even number or a multiple of 3? (1 point) \\(\frac{1}{3}\\) \\(\frac{2}{3}\\) 17 \\(\frac{10}{9}\\)

Explanation:

Step1: Calculate the number of even numbers

The even numbers from 1 to 21 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. So \(n(\text{even}) = 10\).

Step2: Calculate the number of multiples of 3

The multiples of 3 from 1 to 21 are 3, 6, 9, 12, 15, 18, 21. So \(n(\text{multiple of }3)=7\).

Step3: Calculate the number of numbers that are both even and multiple of 3

The numbers that are both even and multiple of 3 (i.e., multiples of 6) from 1 to 21 are 6, 12, 18. So \(n(\text{even and multiple of }3) = 3\).

Step4: Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

By the formula \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), we have \(n(\text{even or multiple of }3)=10 + 7-3=14\).
Since the total number of regions \(n(\text{total}) = 21\), the probability \(P=\frac{n(\text{even or multiple of }3)}{n(\text{total})}=\frac{14}{21}=\frac{2}{3}\).

Answer:

\(\frac{2}{3}\)