SOLUTION: I am trying to calculate a probability problem. It reads: "If someone spins the spinner twice, what is the probability it will land on an even number?" The spinner has the follo

Question 1142724: I am trying to calculate a probability problem. It reads: "If someone spins the spinner twice, what is the probability it will land on an even number?"
The spinner has the following numbers clockwise: 1,2,3,2,5,4,4. However, the space that includes the 1 is twice the size of the others and is 25% of the total area of the spinner. If I draw a dividing line in the larger space, I will then have 8 equal parts. Is this the correct procedure for solving the problem?
Found 2 solutions by greenestamps, Theo:
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

That gives you 8 spaces numbered 1, 1, 2, 3, 2, 5, 4, 4, all with equal probability of landing on them.

So that looks like ONE very good way to start working on the problem; then finishing should be easy:
number of even-numbered sections P(even) = ---------------------------------- total number of sections

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added after seeing the response from the other tutor....

The question you asked was whether dividing the large region into two regions each the size of the other 6 regions was a good idea. I replied that it was.

When you continue working on the problem, there is a problem. It is not clear what it means to ask for the probability that if the spinner is spun twice it lands on an even number.

Does that mean land on an even number both times? or exactly one time? or at least one time?

The answers to those three different questions are different....
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.

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