When dealing with probability problems, you must set up a fraction. In the numerator, or the top part of the fraction, you put the number of whatever is being asked for, and in the denominator, or the bottom part of the fraction, you put the total number of items in the sample. 

In this case, you are being asked for the probability of selecting a blue marble. Since there are  blue marbles,  goes in the numerator of your fraction. Since there are  blue marbles and  red marbles, there are a total of  marbles in the bag. So  is the total number of items in your sample, and  is the number that goes in the denominator of your fraction.

Therefore, your answer is .

First, find the probability of drawing a red marble.

Recall what a probability is: 

Since there are  red marbles and a total of , the probability of drawing a red marble is .

Since we are not replacing the marbles, after we've drawn one marble, the total number of marbles left will be .

Next, find the probability of drawing a blue marble. Since there are  blue marbles and a total of  marbles, the probability of drawing a blue marble is then .

Since the question wants the probability of both events happening, we will need to multiply the two probabilities.

The events are NOT mutually exclusive, meaning there is some overlap. That is, there exist cards that are both black AND an ace. Therefore, we have...

P(Ace or Black) = P(Ace) + P(Black) - P(Ace and Black)

In a deck there are 4 aces, 26 black cards, and 2 cards which are both black and an ace.

P(Ace or Black) = 

You choose a lollipop, and then put it aside before choosing again. This is probability WITHOUT replacement. The probability of choosing a grape on the first pick is  because there are 10 grapes out of 15 total. When the second pick is made, one lollipop has been discarded, so there are now 14 left. The probability of getting an orange lollipop on the second pick is . The probability of the two events occurring is 

There are two outcomes when tossing a coin. When tossing 4 coins, there are 2 x 2 x 2 x 2 = 16 possible outcomes by the counting principle. Each outcome is equally likely.

It must be noted, however, that more than one outcome fits the CRITERIA for coming up with two heads. For example, there is THHT, TTHH, HHTT...

The number of outcomes that fit the criteria can be found by taking the number of combinations here. In this case, it is . 

Therefore, there are 6 favorable outcomes out of 16 possible outcomes. The probability of coming up with two heads is  or .

The idea of an event occurring AT MOST 3 times, means that the event occurring 0 times, 1 time, 2 times, or 3 times fit the criteria. So, one has to consider all of those possibilities.

Using binomial distribution, we can assess the probability for each event occurring. 

P(0 tails) = 

P(1 tail) = 

P(2 tails) = 

P(3 tails) = 

P(0 tails or 1 tail or 2 tails or 3 tails) = P(0 tails) + P(1 tail) + P(2 tails) + P(3 tails) [since they are mutually exclusive] = .650

Each outcome on a roll of a regular die is an equally-likely outcome.

In the case of equally-likely outcomes, one can find the probability of an event occurring by dividing the number of favorable outcomes by the number of possible outcomes. The favorable outcomes are 1, 2, 3, 4, and 5. The possible outcomes are 1, 2, 3, 4, 5, and 6. There are 5 favorable, and 6 possible. So...

The probabilities of rolling a die and flipping a coin are independent. The probability of rolling a die and getting an even number is  or  and the probability of flipping a coin and getting tails is .

This question asks to select a marble WITH REPLACEMENT. These are independent events. The outcome of one does not affect the outcome of the other. We have P(blue and green) = P(blue) x P(green). The probability of selecting a green marble is  or  and the probability of selecting a blue marble is  or . So, the probability of selecting one green marble and one blue marble is =

You can select red and yellow in one of two ways. That is  OR .

It is key, in this problem, to recall that the selection is WITH replacement.

If we add the two probabilities, we get 

.