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?

Answer 1

Answer:

$\frac{2}{3}$.        

Step-by-step explanation:        

We are told that a spinner has equal regions numbered 1 through 21.  

We can see that our both events are not mutually exclusive events, so will use formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ to find our desired probability.

$P(A\cup B)$ = Probability of event A or event B.

$P(A)$ = Probability of event A.

$P(B)$ = Probability of event B.

$P(A\cap B)$ = Probability of event A and event B.      

Let us find probability that spinner will stop on an even number. Even numbers on spinner's region are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.  

$P(\text{ Spinner will stop on an even number})=\frac{10}{21}$

Now let us find probability that spinner will stop on a multiple of 3. Multiples of 3 on the spinner's region are: 3,6,9,12,15,18,21.

$P(\text{ Spinner will stop on a multiple of 3})=\frac{7}{21}$

Now we have to figure out probability that spinner will stop on an even number and multiple of 3. Both events happening at one time are: 6,12,18.

$P(\text{Spinner will stop on an even number and multiple of 3})=\frac{3}{21}$

Now let us substitute our values in above mentioned formula.      

$P(\text{Spinner will stop on an even number or multiple of 3})=\frac{10}{21}+\frac{7}{21}-\frac{3}{21}$

$P(\text{Spinner will stop on an even number or multiple of 3})=\frac{17}{21}-\frac{3}{21}$

$P(\text{Spinner will stop on an even number or multiple of 3})=\frac{14}{21}=\frac{2}{3}$

Therefore, probability that the spinner will stop on an even number or a multiple of 3 is $\frac{2}{3}$.