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? (Points : 5)
Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
P(even) = 10/21
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21.
Probability of either/or = sum of individual probabilities - common probabilities.
To find the probability that the spinner will stop on an even number or a multiple of 3, we need to determine the total number of favorable outcomes and the total number of possible outcomes.
Total Number of Even Numbers: Since the spinner has equal regions numbered 1 through 21, half of these numbers will be even. So, there are 21/2 = 10.5 even numbers. However, since we cannot have half a number on the spinner, we round it down to the nearest whole number. Therefore, there are 10 even numbers.
Total Number of Multiples of 3: We need to find how many integers from 1 to 21 are divisible by 3. Dividing 21 by 3, we get 7. So, there are 7 multiples of 3.
Total Number of Favorable Outcomes: Some numbers are both even and multiples of 3. To avoid counting these numbers twice, we need to subtract the number of numbers that fit both criteria (i.e., the multiples of 6). Dividing 21 by 6, we get 3.5, so there are 3 multiples of 6.
Therefore, the total number of favorable outcomes is 10 + 7 - 3 = 14.
Total Number of Possible Outcomes: Since the spinner has equal regions numbered 1 through 21, there are 21 possible outcomes.
Probability = (Number of Favorable Outcomes) / (Number of Possible Outcomes)
Probability = 14 / 21
Probability = 2/3
So, the probability that the spinner will stop on an even number or a multiple of 3 is 2/3.
To find the probability that the spinner will stop on an even number or a multiple of 3, we need to first determine how many outcomes in the sample space satisfy this condition, and then divide this number by the total number of possible outcomes.
We know that the spinner has equal regions numbered 1 through 21.
To find the even numbers, we need to identify the numbers divisible by 2. We can do this by starting with the first even number, 2, and adding 2 repeatedly until we reach the maximum possible number in our sample space, 21.
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
So there are 10 even numbers in our sample space.
To find the multiples of 3, we need to identify the numbers divisible by 3. We can do this by starting with the first multiple of 3, which is 3 itself, and adding 3 repeatedly until we reach the maximum possible number in our sample space, 21.
3, 6, 9, 12, 15, 18, 21
So there are 7 numbers that are multiples of 3 in our sample space.
However, we may have counted some numbers twice. For example, 6 is both an even number and a multiple of 3.
To account for this, we need to find the numbers that are both even and multiples of 3 in our sample space. We can do this by finding the common multiples of 2 and 3, which are multiples of their least common multiple, 6.
6, 12, 18
So there are 3 numbers that are both even and multiples of 3 in our sample space.
To find the total number of outcomes, we just need to count the numbers from 1 to 21, which is 21 outcomes.
Now, we can finally calculate the probability:
Probability = (Number of desirable outcomes) / (Total number of possible outcomes)
Number of desirable outcomes = Count of even numbers + Count of multiples of 3 - Count of even and multiples of 3
Number of desirable outcomes = 10 + 7 - 3 = 14
Total number of possible outcomes = 21
Probability = 14 / 21
Hence, the probability that the spinner will stop on an even number or a multiple of 3 is 14/21 or approximately 0.67.