Q: What is the probability of not spinning a 4 on a spinner with ten spaces labeled 1-10?

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depends on how many 4's and how many total spaces, assuming they are all equal size. If not then it depends on area of each sector.

That depends on what nine numbers are on the spinner! However, if your spinner is numbered 1-9 and the spaces are all the same size, then your chances for an even number are four out of nine (2,4,6,8 are even.)

Discrete probability. It helps if the all the outcomes in the sample space are equally probable but that is not a necessity.

Suppose the moves are based on throws of a regular die. A move ahead of 4 spaces can be achieved by for moves of 1 each, two of 1 and one of 2, one of 1 and one of 3, two of 2 or one of 4. So the probability of these outcomes is (1/6)4 + 3*(1/6)3 + 2*(1/6)2 + (1/6)2 + (1/6) = 0.2647 approx.

2 spaces!

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depends on how many 4's and how many total spaces, assuming they are all equal size. If not then it depends on area of each sector.

all that matters in this problem, is that 2 of the 13 spaces are orange. On the first spin, the odds are 11:2 against Jake getting the orange space. On the second spin, the result of the first spin has no relevance or influence at all, so the odds are still 11:2 against. Why is this question flagging up in "breakups"?

That depends on what nine numbers are on the spinner! However, if your spinner is numbered 1-9 and the spaces are all the same size, then your chances for an even number are four out of nine (2,4,6,8 are even.)

Theoretical probability.

Classical Probability!

Discrete probability. It helps if the all the outcomes in the sample space are equally probable but that is not a necessity.

A player starts the Monopoly Roller Coaster game by rolling the dice or spinning the spinner, depending on the version they have. Then the player moves his piece the number of spaces shown in clockwise direction around the board. The player must act according to the space he landed at.

-5

Empty space - there is a hypothetical probability that the spaces between stellar objects can be filled with "dark matter".

Demetrios A. Kappos has written: 'Probability algebras and stochastic spaces' -- subject(s): Boolean Algebra, Probabilities

A Venn Diagram is set-up by drawing two large interlocking circles. This creates three spaces, the middle space being part of both circles. The right and left spaces are labeled with two terms, and the middle space is used for items that apply to both terms. Items are then placed (written down) into the correct spaces depending on how they apply.

252/1024 or 0.246. One method of calculating it is this: The total number of outcomes possible by tossing a coin 10 times is 2 to the 10th, which is 1024. In addition, getting 5 heads in 10 tosses is like arranging 5 identical objects in 10 spaces (the remaining 5 spaces are by default Tails), which can be done in 10C5 ways, which is 252. So the probability of getting 5 heads is 252/1024.