Probability

The odds are 1 in 5 that the ball will be green.

The likelihood of an event occurring is known as the probability of occurrence. This can be calculated based on previous patterns and other factors.

The probability of getting five heads out of 10 tosses is the same as the probablity of getting five tales out of ten tosses. One. It will happen. When this happens, you will get zero information. In other words, this is the expected result.

Imagine that the dice are red one and a blue one. There are two answers to this. The first is 36 - this comes from the fact that the red die can give any result from 1 to 6, and those six numbers can each be matched with the six number which the blue die can give. However, to most people it doesn't matter if the red die gives 4 and the blue one gives three, - they see that as the same result as if the red die gave 3 and the blue one gave 4. Similarly to many people a red 5 and a blue 2 would be the same 'result' as a red 6 and a blue 1. Looked at this way there are 11 different 'results' which two dice can give - ranging from 2 to 12. So, the 'permutations' number 36 and the 'combinations' number 11.

450,695

The value of a probability is a number between 0 and 1

So it's either positive and less or equal to one or null

Best way to answer this question is by figuring out the probability of a phone number not containing the number 7. There are 9 other numbers that are not a 7, so odds of each digit not being a 7 is 90%. For a 7-digit number, calculation would be 0.9*0.9*0.9*0.9*0.9*0.9*0.9 or 0.9^7. This equals 47.8%, which means the odds of a phone number containing at least one 7 is 100% - 47.8%, which is 52.2%. But phone numbers do not begin with 0 or 1, so if we account for that, the odds of the first digit not being a 7 is 7/8, not 9/10. We replace the first 0.9 with 0.875, and the result is 53.5%. There are other scenarios we can take out, like numbers that begin with 555, but I'll just leave it at that.

The probability of a woman being chosen in this situation is 7/11.

The first settlement on American soil was in 1565 when the Spanish settled in what is now St. Augustine, FL.

In stratified sampling, the population to be sampled is divided into groups (strata), and then a simple random sample from each strata is selected. For example, a state could be separated into counties, a school could be separated into grades. These would be the 'strata'.

1/36

A probability measure allocates a non-negative probability to each possible outcome. All individual probabilities together add up to 1.

The "risk-neutral probability measure" is used in mathematical finance. Generally, risk-neutral probabilities are used for the arbitrage-free pricing of assets for which replication strategies exist. This is about relative pricing, based on possible replication strategies. The first argument is that a complete and arbitrage-free market setting is characterised by unique state prices. A state price is the price of a security which has a payoff of 1 unit only if a particular state is reached (these securities are called Arrow securities). In a complete market, every conceivable Arrow security can be traded. It is more easy to visualise these securities in terms of discrete scenarios. (On a continuous range of scenarios we would have to argue in terms of state price density.) The arbitrage-free price of every asset is the sum (over all scenarios) of the scenario-payoff weighted with its state price. Any pricing discrepancy with regards to an implicit state price would enable arbitrage in a complete market. The assumption is that the pursuit of such opportunities drives the prices towards the arbitrage-free levels. Hence the state prices are unique.

Since the whole set of Arrow securities is the same as a risk-free bond (sure payoff of 1 unit at maturity), the price of the whole set of Arrow securities must be e^(-rt) (assuming we are now at maturity minus t). Risk-neutral probabilities can then be defined in terms of state prices, or vice versa. A probability measure has to fulfil the condition that the sum of all individual probabilities adds up to 1. Therefore, if we want to create an artificial probability distribution based on the state price distribution, we have to multiply each state price with e^(rt) in order to obtain its probability equivalent. It is not surprising then that any expectation taken under the risk-neutral probability measure grows at the risk-free rate. This is an artificial probability measure, why should we create such a construct? This connection allows us to exploit mathematical tools in probability theory for the purpose of arbitrage-free pricing. The main difficulty about risk-neutral probabilities is that the probability concepts used have not initially been developped for the purpose of financial pricing, therefore, two different languages are used, which can easily be confusing. The economic interpretation of a risk-neutral probability is a state price compounded at the risk-free rate. Anything that has an effect on a state price (preferences, real probability, ...), has an effect on the risk-neutral probability.

So now we have a bridge to go from state prices to risk-neutral probabilities and back again. What is this good for? According to the second argument, we can, under certain conditions, specify the unique risk-neutral probability distribution of an underlying asset price with the help of an only incomplete specification of its real probability distribution, thanks to the Girsanov Theorem. If the innovation in the price of the underlying asset is driven by a Brownian motion, then all we need to obtain the risk-neutral probability distribution is the volatility parameter. What can we now do with this risk-neutral probability distribution? We can use the first argument to convert the obtained risk-neutral probability distribution back to a state price distribution, and the state price distribution applied to the payoff distribution (i.e. taking the sum over all scenarios) leads to the arbitrage-free price. These arguments save us a lot of trouble when trying to calculate the arbitrage-free price of an asset. They allow us to avoid the estimation of risk premia, by implicitly using those incorporated in the underlying asset price. The arbitrage-free price is, however, NOT independent of risk-premia. The price of the underlying asset is part of the pricing equation, and the risk-premia are inherent in this price, but because the price of the underlying asset is known to us, we obviously do not need estimate it.

It is important to emphasise that the risk-neutral valuation approach only works if the asset to be priced can be perfectly replicated. This is often not true in reality, especially when dynamic replication strategies are involved.

Paper explaining risk-neutral probabilities:

http://ssrn.com/abstract=1395390

It is 156/2652 = 0.0588

Almost 12 %

It's the same as flip a penny ten times.

There are 2^10 = 1024 different outcomes (each penny being either head or tail).

The ones we can 'use' are those with 3 heads.

Let '1' be head and '0' be tail:

0001010100 is a valid outcome - and 1110000000 as well.

So how many way can we arrange three 1's in ten spaces?

In 120 ways.

120 ways out of 1024 is the answer.

120/1024=0.1171875 = 11.71875 %

5/18

It doesn't matter what number the first roll produces. The chance that the second throw will be the same is 1/6, so the chance of throwing one of the five different remaining numbers with the second dice is 5/6. Now there are only 4/6 ways of scoring with the third throw, and finally only 3/6 possibilities for the last throw.

So the probabiliy of throwing 4 different numbers = 1 * 5/6 * 4/6 * 3/6, w

hich is 5/6 * 2/3 * 1/2 = 5/18 = 0.2777777777777777or 27.77777777%

which is almost 28%

The probability based on an event that has already occurred is 100%. If the event has occurred, it has occurred.

it is a ten sided and ten angled figure

The term "theoretical probability" is used in contrast to the term "experimental probability" to describe what the result of some trial or event should be based on math, versus what it actually is, based on running a simulation or actually performing the task. For example, the theoretical probability that a single standard coin flip results in heads is 1/2. The experimental probability in a single flip would be 1 if it returned heads, or 0 if it returned tails, since the experimental probability only counts what actually happened.

No because it isn't one of the normal counting numbers.

Each of the dice has 6 sides.

There are 6 x 6 = 36 possible combinations.

Of those, the following equal 5:

1 and 4

2 and 3

3 and 2

4 and 1

So the odds of rolling a 5 is

4 chances out of 36

4 / 36 = 1 / 9 That converts to a probability of 0.1111111...

4 7's

1/26

9/2

Measurements of the variable of interest will differ between subjects. As a result the outcome of an experiment may be different from what was expected. The psychologist will want to know whether that difference is purely the result of random variation between subjects or if it is due to some other factor. The standard deviation is a measure for determining how large a difference might be purely random.