Aspinner 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?

Using it's concept, it is found that there is a 0.6667 = 66.67% probability that the spinner will stop on an even number or a multiple of 3.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, there are 21 possible outcomes for the spinner.

Hence:

0.6667 = 66.67% probability that the spinner will stop on an even number or a multiple of 3.

More can be learned about the probability concept at link

.        

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 to find our desired probability.

= Probability of event A or event B.

= Probability of event A.

= Probability of event 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.  

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.

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.

Now let us substitute our values in above mentioned formula.      

Therefore, probability that the spinner will stop on an even number or a multiple of 3 is .    

asdsaddsa

step-by-step explanation: