In order to solve this problem, we need to discuss probabilities. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

The die has six sides with the following values: one tow, three, four, five, and six. Of these values the die has three that are even: two, four, and six. We can write the following probability.

Simplify.

Now, let's convert this into a percentage:

Two fair dice can be thrown with 36 equally probable rolls. 18 rolls comprise one even number and one odd number. They are as follows - with the rolls that total seven in boldface:

6 of the 18 possible rolls have total 7, making the probability of a roll of 7, given that exactly one doe shows an even number, equal to  .

Two fair dice can be thrown with 36 equally probable rolls.9 rolls can be thrown that show two odd numbers: they are

This leaves 27 rolls with at least one even number, 6 of which result in a sum of 7:

This makes the probability of rolling a sum of 7

,

the correct choice.

In order to solve this standard we need to understand two primary components: probabilities and the property of independent events. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

An intersection is the likelihood that one event will occur with another.

We can calculate that the probability of rolling a one on a six-sided die is:

Now, let's determine the probability of flipping a coin to reveal the "heads" side. A coin has two sides: heads or tails. The probability of rolling heads is as follows:

An intersection is the probability of two events occurring simultaneously; therefore, we need to multiply the probabilities.

Convert to a percentage.

Step 1: If we have 3 quarters and each quarter has two outcomes, then there are  total outcomes that can come out of all of the rolls of the quarters. 

Those rolls are:

HHH, HHT, HTH, THH, THT, TTH, TTT, HTT.

Step 2: Determine how many outcomes have no more than  heads....

We cannot count HHH because it has  heads. 

We also cannot count TTT because it has no heads at all.

Step 3: We had  options and we can't take  options..so we have a total of  options left. 

Step 1: Find the probability of getting a green marble...

The probability of a green marble is .

Step 2: Find the probability of getting a blue marble (without replacement)

When we have without replacement, we do not put the first marble we take out back into the sack. In this case, the total number of marbles in the sack is  less.

Probability of a blue marble is .

Step 3: To find the probability of both events happening at the same time, we need to multiply the probabilities in the first two steps.

So, 

Step 4: Simplify...

 is the answer. 

The number of ways to choose four people to sit at the first table without regard to the arrangement at either table is the number of combinations of four people from a set of eight; once the people who will sit at the first table are chosen, the people who will sit at the second are automatically chosen. Therefore, the correct number of such arrangements is .

Since

,

then, setting ,

,

the correct number of arrangements.

In a deck of 52 cards, one-fourth of the cards, or 13 cards, are spades. Also, there are four aces, one of which is a spade and three of which are not spades. It follows that the removal of the spades and the aces results in the removal of

cards, to leave a modified deck of 

cards.

None of the two red threes (the three of clubs, the three of diamonds) were removed, so the probability of a random draw of a red three is

.

Because the last test cannot be either math or writing, the selection of an order in which the tests are taken must be looked at as a sequence of events as follows:

First, the test to be taken last must be one of three tests: reading, science, or social studies;

Second, the remaining four tests must be arranged in a particular order. There are four ways to schedule the first test, three remaining ways to schedule the third, two for the third, and one for the fourth.

By the multiplication principle, this makes

possible schedules.