Question 1142724
the probability that the spinner will land anywhere on the spinner board is considered to be equal.


since the number 1 space is 25% of the area of the board, then the other numbers, being equally spaced, represent 75% of the area of the board.


you have 6 other numbers.
75% / 6 = 12.5


each of the other numbers takes up 12.5% of the area of the board.
if you split the 1 into 2 spaces, then each one of the number 1's will also take up 12.5% of the board.


it sounds like your method will work.


the numbers on the board are then 1,1,2,3,2,5,4,4.


each of those numbers takes up 12.5% of the area of the board.


4 of the numbers are even and 4 of the numbers are odd.


the probability that the spinner will land on an even number is therefore 4 / 8 = 50%.


the probability that the spinner will land on an odd number is therefore also 4 / 8 = 50%.


on your first spin, the probability that it will land on an even number is .5.


on your second spin, the probability that it will land on an even number is also .5.


the probability that it will land on an even number both times is therefore .5 * .5 = .25.


there is another way to look at it.


the number 1 takes up 25% of the board space.
the number 3 takes up 12.5% of the board space.
the number 5 takes up 12.5% of the board space.


all the odd numbers take up 50% of the board.
all the even numbers take up 50% of the board.
same probabilities.