a spinner has regions numbered through 1-21
what is the probability that the spinner will stop on an even number or a multiple of 3..
I am confused.
I keep coming up with 17/21
but the real answer is 2/3
how how how ?
A= 16/21
You need to remember that 6 is also an even number, so you can't count it twice as it is only one number. So you have 10 possible even numbers, plus 7 possible numbers that are multiples of 3, minus 1 multiple of 3 (the number 6) because you already counted it in your even numbers.
Even number = 10/21
Multiple of 3 = 7/21
10/21 + 7/21 = 17/21
Guess what? I keep coming up with 17/21 too.
Book typo?
A spinner has 5 sections labeled 1 through 5. The probability of randomly spinning an even number is 2/5. What is the probability of not spinning an even number?
To calculate the probability that the spinner will stop on an even number or a multiple of 3, you need to determine the number of favorable outcomes (even numbers or multiples of 3) and divide it by the total number of possible outcomes (numbers 1-21). Let's break down the problem into steps:
Step 1: Determine the favorable outcomes.
- Even numbers: The even numbers between 1 and 21 are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20. So, there are 10 even numbers.
- Multiples of 3: The multiples of 3 between 1 and 21 are 3, 6, 9, 12, 15, 18, and 21. There are 7 multiples of 3.
- However, since the number 6 appears in both lists, we need to count it only once to avoid duplication. So, the total number of favorable outcomes is 10 + 7 - 1 = 16.
Step 2: Determine the total number of possible outcomes.
The spinner has regions numbered 1 through 21, so there are 21 possible outcomes.
Step 3: Calculate the probability.
Divide the number of favorable outcomes (16) by the total number of possible outcomes (21) to get:
Probability = 16/21.
Hence, the correct answer is 16/21, not 17/21. It seems there might have been a calculation error in obtaining the real answer of 2/3.