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.

的refore, investigate when die one is three.

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

的refore 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.

的probability thattrials will result insuccesses in a Bernoulli process is

,

whereis the probability of a success in any given trial andis 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 probabilitythat exactlyheads will come up is

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

Of the five choices, 0.25 comes closest.

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

的easiest way to find the probability that Jan will complete her large straight - by rolling a "5" - is to find the probability that she willnotcomplete it. This happens if she rolls any number other than "5" on both rolls, which, by the multiplication principle, will happen with probability

的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.

的se 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.

的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.

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

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

的eventis 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,

的probability of the event is

.

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

的eventis the set of all combinations of two cards from the set of four red cards. The size of this event space is

的probability of the event is