Question

a spinner is divided into eight equal parts numbered 1 to 8. if the spinner is spun once, what is the probability that it stops on a number that is a multiple of 2?

Answer 1

Answer:

$\frac{1}{2}$, which is 50% probability.

Step-by-step explanation:

The spinner equally divided into 8 parts and named as 1, 2, 3, 4, 5, 6, 7 and 8.

It is spun once, we need to find the probability of getting a number that is a multiple of 2.

Let's find the multiples of 2: 2, 4, 6, and 8

The total number favorable outcomes = 4

The number of possible outcomes = 8

The probability of an event = The number of favorable outcomes / The total number of possible outcomes.

Therefore, the probability  that it stops on a number that is a multiple of 2 = $\frac{4}{8}$

When we simplify this fraction, we get

= $\frac{1}{2}$

Which is 50% probability.

Answer 2

Since the total number of available selections is divided by two, the probability is 50%.  If we look at the total set of numbers, we find that even numbers are multiples of 2.  2, 4, 6, 8 (which makes up half of the numbers on the spinner).