We will consider the value of each event to Jill; positive values refer to Jack paying Jill, and negative values refer to Jill paying Jack.

There are 53 equiprobable outcomes. Along with their probabilities and their values to Jill, they are:

: The joker is drawn. 1 out of 53 outcomes is favorable to this event, so

The value to Jill, , is what is to be determned.

: A club is drawn. 13 out of 53 outcomes are favorable to this event, so

Since Jill wins $100 in this event, the value to her is

: A card other than a club or a joker is drawn. 39 out of 53 outcomes are favorable to this event, so

Since Jill loses $20, the value to her is 

.

The expected value of any game can be calculated by multiplying the probability of each event by its value, and adding the products. That is,

The game is to be fair, meaning that . Making this and the other substitutions, we get

Solve for

Jill must agree to pay Jack $520 in the event of the draw of the joker.

This is a conditional probability problem.

There are thirteen spades and thirteen clubs; since the first card is known to be black, the probability of the first card being a spade (or club) is

,

The probability of the second card being a spade depends on the suit of the first card. The probability of the second card being a spade, given that the first card was also a spade, is

The probability of the second card being a spade, given that the first card was a club, is

The overall probability that the second card will be a spade is

First, recall that each die is rolled independently.

Now, it is known that the first die is an odd number therefore the possible options are:

From here, find the number of combinations each one will have to make a sum of 8 with the other die.

By investigation, die number one cannot be one because it along with any value from die two will not equal eight.

Therefore, investigate when die one is three.

Next, if die one is five then the possible sum is,

Therefore there are a total of two possibilities of rolling a sum equaled to eight.

From here, identify the total number of options the sum could be when it does not equal eight.

Since each trial can have one of two outcomes, heads or tails, this is an example of a Bernoulli process.

The probability that trials will result in  successes in a Bernoulli process is

,

where is the probability of a success in any given trial and is the probability of a failure.

Since there are six tosses, set . We can call heads the success with probability , and tails the failure with probability . Accordingly, the probability that exactly heads will come up is

Since we are looking for the probability of heads coming up at least five times, evaluate as follows:

Of the five choices, 0.25 comes closest.

Let be the event that Jan gets a straight - that is, that she rolls a "5" on her second or third turn.

The easiest way to find the probability that Jan will complete her large straight - by rolling a "5" - is to find the probability that she will not complete it. This happens if she rolls any number other than "5" on both rolls, which, by the multiplication principle, will happen with probability

The probability that she will get the straight is this probability subtracted from 1, or

Suppose the coin is indeed fair. The probability of the coin coming up heads in one trial is ; by the multiplication principle, the probability that the coin will come up heads each of ten consecutive times is . While this makes this event extremely unlikely, it is not impossible. Thus, the fact that the coin comes up ten heads in ten flips is not conclusive evidence that the coin is unfair.

In order for Mike to get his five-of-a-kind, one of three things has to happen:

1) He can roll two "3's" on his next turn. There is only one way out of thirty-six for this to happen, so the probability of rolling two 3's is .

2) He can roll one "3" on his next turn, then one "3" on his third turn. There are ten ways out of thirty-six to roll exactly one "3" with a pair of dice (1-3, 2-3, 4, 3, 5-3, 6-3, 3-1, 3-2, 3-4, 3-5, 3-6), making this probability . After this, he will roll one die; the probability of rolling a "3" is . By the multiplication principle, the probability of this scenario happening is .

3) He can roll no "3's" on his next turn, then two "3's" on his third turn. There are twenty-five ways out of thirty-six to roll two dice and not get a 3, so the probability of this is ; the probability of rolling two "3's" on the third turn is . By the multiplication principle, the probability of this scenario happening is .

These three scenarios are mutually exclusive and together comprise the event that Mike gets his two 3's. Add the probabilities:

Empirical probability is based on repeated experimentation and observation; theoretical probability is calculated based on equally probable events and the number of such events favorable to a given outcome.

The probability that a black ball will be drawn is hypothesized based on an experiment, in which the number of times a black ball was drawn compared to the number of total draws was observed. The probability was derived based on this observation. This is therefore an example of empirical probability.

Empirical probability is based on repeated experimentation and observation; theoretical probability is calculated based on equally probable events and the number of such events favorable to a given outcome.

The two outcomes for the flip of a fair coin are two, which are equally probable. The probability of each outcome is based on theoretical probability.

Two cards are drawn from the deck without regard to order, so the sample space  is the set of all combinations of two cards from a set of twelve. The size of this sample space is 

The event is the set of all selections of one card from a set of six red cards and one card from  a set of six black cards. The selections are independent, so by the multiplication principle,

The probability of the event is

.