While the fundamental counting principle, permutations, and combinations deal with all possible combinations that can occur, probability deals with the likelihood of a specific event occurring. For example, you know that when you toss a penny it is just as likely the penny will land heads up as it will land tails up. The penny has two sides, and each side is considered one part of the penny, or one half. Since both sides of the penny are one half of the penny (ignoring the rim of the penny), the probability of the penny landing heads up is the same as landing tails up.

The following is the probability of an event, *A*, happening:

The probability of an event happening is between 0 and 1, inclusive. If an event has a probability of 0, it will never happen, and if an event has a probability of 1, it will definitely happen.

**For Example**

A box contains 3 red marbles, 4 blue marbles, and 5 white marbles. What is the probability of drawing a blue marble from the box?

Probability is similar to ratios. The probability of drawing a blue marble is the same as the ratio of blue marbles to total marbles.

Add up the total number of marbles

(3 + 4 + 5 = 12)

The ratio of blue marbles to total marbles is 4:12, or 1:3 reduced.

What is the probability of drawing a green marble? Since the box contains no green marbles, the probability would be 0/12.

**Independent Events**

*P*(*A*) × *P*(*B*) = *P*(*I*)

Events A and B are independent if the outcome of one has no effect on the outcome of the other.

For example, a bag contains 3 red marbles and 4 blue marbles. If you draw a blue marble from the bag and then replace it, what is the probability of drawing two blue marbles in a row?

Since you are replacing the blue marble, the probabilities of drawing a blue marble each time are the same:

**Dependent Events**

*P*(*A*_{1} ) × *P*(*A*_{2} ) = *P*(*D*)

Events A_{1} and A_{2} are dependent if the outcome of A_{1} affects the outcome of A_{2}.

For example, a bag contains 3 red marbles and 4 blue marbles. If you draw a blue marble from the bag and don’t replace it, what is the probability of drawing two blue marbles in a row?

Since you are NOT replacing the blue marble, the second probability assumes you drew a blue marble the first time. Since you didn’t replace it, the bag now has 3 blue marbles and 6 total marbles:

Mutually exclusive events are similar to independent events in that the outcome of one event has no impact on the outcome of the second event. The difference is that mutually exclusive events refer to a single occurrence, while independent events refer to several occurrences. A safe indicator for independent events is the word “and” placed between events, while a safe indicator for mutually exclusive events is the word “or” placed between events.

For example, in a single toss of a six‐sided die numbered 1 to 6, I can roll a 1, 2, 3, 4, 5, or a 6. The likelihood of rolling one of those numbers is one‐sixth. The likelihood of rolling a 1 or a 2 is greater than the probability of rolling just one of the numbers. We add the probabilities together to reach the final probability of mutually exclusive events.

Mutually exclusive events are events that cannot happen at the same time.

*P*(*A*) + *P*(*B*) = *P*(*M*)

Examples of mutually exclusive events include: flipping heads and tails on a coin, drawing a jack and a king from a deck of cards.

Mutually exclusive events are defined by *P*(*A *and *B*) = 0*.*

**Symbols **

- Another way of writing
*P*(*A*and*B*) is*P*(*A*∩*B*). - Another way of writing
*P*(*A*or*B*) is*P*(*A*∪*B*).

Imagine a Venn Diagram:

*A *∪ *B* is anything contained in either A or B. 𝐴∩𝐵 is anything contained in both A and B (the overlapping section).

For mutually exclusive events, *P*(*A *∩ *B*) = 0. Therefore, a Venn Diagram for mutually exclusive events would look like this:

**Example Problem**

What is the probability of rolling a 1 or a 2 on a six sided die numbered 1-6?

Landing a 1 or a 2 are mutually exclusive events, since they cannot happen at the same time.

*Event A *= rolling a 1, *Event B* = rolling a 2

*P*(*A*) + *P*(*B*) = *P*(*M*)

We can also write the probability of a mutually exclusive event as *P*(*A *∪ *B*) = *P*(*A*) + *P*(*B*).

The roll can either land on 1 or 2, but not both. By adding the two probabilities together, we calculate the likelihood of rolling either number.

Not mutually exclusive events can occur at the same time. Examples of non mutually exclusive events include: drawing a king and a heart from a deck of cards, taking Latin and History classes, rolling a 5 on a die and drawing a 5 from a deck of cards.

Not mutually exclusive events can occur at the same time.

To find the probability of *A* or *B*, we need to add together the probability of *A* and the probability of *B*, just like we did for mutually exclusive events. This time, however, we must also subtract the probability of *A* and *B* occurring at the same time (the overlapping section), so we don’t double count.

*P*(*A *∪ *B*) = *P*(*A*) + *P*(*B*) - *P*(*A *∩ *B*)

For Eample, what is the probability of drawing a jack and drawing a spade from a standard deck of cards?

*Event A* = drawing a jack; *Event B* = drawing a spade

The probability of A and B is the probability of drawing the jack of spades. Since there is only one in a deck, the probability is 1/52.

A factorial is marked by an exclamation point (!), but it does not mean that the number is exciting. When you see an exclamation point, or factorial, next to a number, it means to find the product of the number and all the natural numbers that come before it.

n factorial (𝑛!) is the product of *n* and all of the natural numbers less than *n*.

*n*! = *n*(*n *− 1)(*n *− 2)…(2)(1)

4! = 4 × 3 × 2 × 1 = 24

Factorials can also be expressed as *n*! = *n*(*n *− 1)!

6! = 6 × 5!

6! = 6 × (5 × 4 × 3 × 2 × 1) = 6 × 120 = 720

A combination is an arrangement of different values where the order of them does not matter. This often occurs when we have a group of people going to some event. For instance, if your parents told you to choose 3 friends to go to the movies with, it doesn’t matter if you go with Alice, Bob and Charlie or with Charlie, Alice, and Bob. We consider both of those groups to be the same thing, a combination of people! Of course, we can extend this idea to items as well. We can solve for how many different possible combinations exist given certain parameters.

To find the number of combinations, we can use the formula

*n*is the total number of items in the set*r*is the number of items we are selecting from the set

For exampe, there are 10 people in soccer club, but they can only send a team of 5 to playoffs. How many 5 person teams can be formed?

We are looking for the number of possible team combinations. In this case, the order does not matter, so we can use our combinations formula.

*n*= 10*r*= 5

There are 252 different possible combinations of teams.

Permutations are like combinations, but the order does matter. An example of a permutation would be positions in a school club. Having Alice be the president, Bob be the vice president, and Charlie be the treasurer will be different than having Charlie be the president, Alice be the vice president, and Bob be the treasurer.

When there are distinctions being made due to the order of the items, there will **always** be a greater number of permutations than combinations!

Permutations are combinations where the order matters. To find the number of permutations, we can use the formula:

*n*is the total number of items in the set*r*is the number of items we are selecting from the set

The formula for permutations is very similar to the formula for combinations, but permutations does not have an* r*! on the bottom.

For example, a four digit lock can use digits 0 through 9 in its passcode. How many different possible permutations of the passcode are possible?

Since in the case of a passcode, the order of the numbers does matter, we can use our formula for permutations.

*n*= 10 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)*r*= 4

There are 5040 different possible permutations of the lock.

- A bag of coins has 9 red coins, 15 green coins, and 17 black coins. What is the probability of drawing a red coin and then a green coin if the first coin is not returned?
- What is the probability of landing a tail on a coin flip if you roll a 6 on a die?
- Simplify
- Betty is going on a vacation and needs to pack 5 shirts. If she owns 30 shirts, how many different possible combinations of shirts are there?
- If 5 cards are laid out in a row from a standard deck, how many possible ordered combinations are there?

**Answers to Practice Problems**

- 27/328
- ½
- 1 / (x
^{2 }- 3x + 2) - 142,506
- 311,875,200

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