Each coin toss is an independent event, meaning the outcome of one toss does not affect the outcome of the other coin tosses.  To find the overall probability, multiply the probabilities of each toss together:

Selecting a specific sequence of marbles from the bag without replacement is a dependent or conditional probabiliy. If you successfully draw a blue marble, it affects the chances of subsequently drawing a gold marble.

In the notation of probability we can call the probability of drawing a blue marble  and the probability of drawing a gold marble .

If we replaced the first marble we drew, we would simply multiply  times , but since we aren't we want to know  or, the probability of B, given A.

So the probability of A and B is equal to:

In words, you have a six in ten chance of drawing a blue marble. If you successfully do that, you then have a four in nine chance of drawing a gold marble. Multiply those odds together, and reduce the fraction and you get a four in fifteen probability.

First, calculate the number of students not planning on attending an in state college.

Either subtract those planning on attending an in state college from the total...

50 - 20 = 30

...or add up all the other categories.

12 + 8 + 10 = 30

Then divide by the total number of students.

30/50 = 60%

The probability of flipping the coin in this particular order is calculated by multiplying the probabilities of each turn by each other. Because we have a "loaded" coin the probabilities are as follows:

 .

Then simplify:

 .

This fraction is in its simplest form.

There are 4 aces in a deck of 52 playing cards.

After 1 ace is removed, there are 3 aces and 51 cards remaining:

To determined the probability of both events, multiply the probability of drawing each card:

Playing a sport, which we can call P(A), and owning a dog, which we can call P(B), are traits that we can assume are independent. The probability of one does not affect the probability of the other.

So, P(A and B) = P(A) * P(B)

Fill in the probabilities given into this equation.

P(A and B) = (0.65) * (0.48) = 0.312 = 31.2%

The probability of finding either a blue car (A) or a green car (B) is written
P(A or B).

This type of probability can be found using the equation:

P(A or B) = P(A) + P(B) - P(A and B)

We need to solve this equation for P(B).

Two of the terms are given in the prompt:

P(A or B) = 0.33

P(A) = 0.22

The third, P(A and B) can be inferred. Since a car can be only one color,
P(A and B) = 0

Thus,

0.33 = 0.22 + P(B) - 0

P(B) = 0.11

The probability of drawing a face card P(A) or a red card P(B) can be written as P(A or B), and be calculated using:

P(A or B) = P(A) + P(B) - P(A and B)

The first two terms on the right side of the equation are fairly straightforward. Since we'll be adding and subtracting fractions, I will keep the denominator the same and won't reduce the fractions until the end.

P(A) = 12/52

P(B) = 26/52 = 1/2

The third, P(A and B), can be calculated using:

P(A and B) = P(A) * P(B)

P(A and B) = 12/52 * 1/2 = 12/104 = 6/52

(Conversely, think there are 12 face cards in a deck of 52, half of which [6] are red.)

Back to our original equation!

P(A or B) = 12/52 + 26/52 - 6/52

= 32/52 = 8/13

Binomial probability

There are ten marbles in the bag and there are only two colors. One of the colors is red and there are six of them.

Because it's asking for red, we write a fraction.

The top of the fraction represents what the question is asking and the bottom represents the total possible outcomes.

So our answer is 

.