Next Generation Quantitative Reasoning, Algebra, and Statistics Study Guide for the ACCUPLACER Test

Probability and Sets

Probability is the chance that a particular event will take place.

Calculate Probability

Consider the following situation: if a fair \(6\)-sided die is rolled, what is the probability that a \(1\) is rolled?

To calculate this probability, it is first necessary to determine how many possible ways a \(1\) can be rolled; it is then necessary to determine how many different possible outcomes there are.

Given that it is a \(6\)-sided die, there is only one way to roll a \(1\), and there are \(6\) possible outcomes.

The probability of rolling a \(1\) is then \(\frac{1}{6}\). This is a case of simple probability, which is described algebraically as the ratio of a successful event to the total number of possible events.

Compound probability problems entail a combination of multiple probabilities.

Consider the following example:

A fair coin is flipped \(2\) times and a fair \(6\) sided die is rolled twice. What is the probability that \(2\) heads are flipped and a \(2\) is rolled twice?

The key word, and (rather than or), informs us that the combined probability is found by multiplying the probability of each individual event with each other. The probability of flipping heads is:

\[\frac{1}{2}\]

The probability of doing this twice in a row is:

\[\frac{1}{2} \cdot \frac{1}{2}\]

Likewise, the probability of rolling a \(2\) twice in a row is:

\[\frac{1}{6} \cdot \frac{1}{6}\]

The overall probability is the product of each individual probability:

\[\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{144}\]

Another example:

What is the probability of rolling a \(4\) or a \(5\) with a fair \(12\)-sided die?

In this case, a successful event is rolling either a \(4\) or a \(5\), so the two events’ probabilities can be combined through addition:

\[\frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}\]

Notice that in each of the cases presented thus far, none of the determined probabilities depended upon any of the other probabilities; flipping a heads or tails was always \(\frac{1}{2}\) regardless of any flip before it.

Likewise, the probability of rolling any number on a die was independent of any previous roll. Conditional probability deals with events that depend on other events.

Consider this example:

A bowl contains \(4\) blue marbles, \(3\) red marbles, \(5\) yellow marbles, and \(6\) black marbles. What is the probability of randomly selecting a blue marble?

If the blue marble is not replaced, what is the probability of selecting another blue marble?

To calculate the first probability, compute the ratio of the number of blue marbles to the total number of marbles:

\[\frac{4}{18} = \frac{2}{9}\]

After the blue marble is removed from the bowl, there are \(3\) blue marbles and \(17\) total marbles. Therefore, the probability of selecting a blue marble after one blue marble has been removed is:

\[\frac{3}{17}\]

The second event depends on the first event, as the first event alters the probability of the second. Now we can find the probability of both events occurring (that is, selecting two blue marbles without having replaced the first) by multiplying these together:

\[\frac{2}{9} \cdot \frac{3}{17} = \frac{6}{153}\]

The probability of drawing the two blue marbles in a row without replacing the first is \(\frac{6}{153}\).

Note that this is different from what we would have if we just used the same probability for each time we select a marble:

\[\frac{2}{9} \cdot \frac{2}{9} = \frac{4}{81}\]

is approximately \(0.05\), whereas \(\frac{6}{153}\) is approximately \(0.04\). Drawing a second blue marble without replacing the first is less likely because there are still the same number of other marbles but the proportion of blue marbles has decreased.

Set Notation

A set is a collection of distinct objects. For example, we can define \(X\) to be the set of whole numbers or integers greater than \(0\) and smaller than \(7\), written as \(X = \{1, 2, 3, 4, 5, 6\}\).

Likewise, let’s define \(Y\) to be the even integers larger than \(0\) and smaller than \(10\). Written in set notation, \(Y = \{2, 4, 6, 8\}\).

We may be interested in combining the objects of both sets, without repeating objects that are in both sets. This is defined as the union of sets and is an operation represented by the symbol \(\cup\).

The union of \(X\) and \(Y\) is then:

\[X \cup Y = \{1, 2, 3, 4, 5, 6, 8\}\]

We might also be interested in listing only the objects present in both sets; this is defined as the intersection of two sets and is symbolically described with a \(\cap\) symbol.
The intersection of \(X\) and \(Y\) is then:

\[X \cap Y = \{2, 4, 6\}\]

Sample Spaces

The sample space of a statistical event is the set of all possible outcomes.

When flipping a coin, the sample space includes the possibility of flipping heads and the possibility of flipping tails.
When rolling a \(6\)-sided die, the sample space includes the possible outcomes of rolling \(1, 2, 3, 4, 5, \text{or} \;6\).

Sample spaces are represented using set notation, which is described above.

So, the sample space of a fair coin is:

\[\{heads, tails\}\]

and the sample space of a fair die is:

\[\{1, 2, 3, 4, 5, 6\}\]

Events

An event is the occurrence of a possibility in the sample space.
Flipping heads or tails is an event when determining the probability relating to flipping a coin. Rolling a \(2\) is an event when rolling a \(6\)-sided die.