Probability and Statistics for the DAT

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Learn key DAT concepts related to probability and statistics, plus practice questions and answers

Probability and Statistics for the DAT banner

everything you need to know regarding probability and statistics for the dat

Table of Contents

Part 1: Intro to probability and statistics

Part 2: Probability

a) Basic probability

b) Combinations and permutations

Part 3: Statistics

a) Mean, mode, and median

b) Normal distribution

c) Unions and intersections

Part 4: Helpful terms

Part 5: Questions and answers

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Part 1: Intro to probability and statistics

Probability and statistics related questions are a crucial part of the quantitative reasoning section of the DAT. These topics can show up as normal questions, or may be word problems. Additionally, probability or statistics may be the topic of data sufficiency or quantitative analysis questions. A good understanding of both probability and statistics will set you up for success on your exam. This guide will teach you everything you need to know for the DAT, with practice questions and answers throughout.

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Part 2: Probability

a) Basic probability

Probability is a numerical representation of how likely an event is to occur. Mathematically, probability is represented as follows:

\[P(A) \space=\space \frac{number\space of\space events\space in\space which\space A\space occurs}{number\space of\space events\space that\space can\space occur} \]

A is a certain event
P(A) is the probability that event A will occur

On the DAT, you will see questions that ask for the probability of both single and multiple events occurring. Let’s start with an example of single event probability, then work through a problem involving multiple events.  

A coin purse contains 4 dimes, 7 nickels, and 2 quarters. What is the probability that a randomly selected coin will be a dime? 

To solve this problem, recognize that the event (A) is selecting a dime at random. Start by determining the numerator, which is the number of events in which A occurs. For this question, the numerator is the number of dimes in the coin purse (4). 

Next, determine the denominator. The denominator is the number of total events that can occur. Because only one coin is being selected, the number of total events that can occur is the number of total coins (13). 

Now, plug these numbers into the original equation.

\[P(A)=\frac{4}{13}\]

Therefore, the probability of a randomly selected coin from the purse being a dime is \(\frac{4}{13}\), or about 30%. Single event problems are quite straightforward, but DAT questions can be a bit more complicated. Let’s look at a multi-event probability question.

A sock drawer contains 7 red socks, 12 white socks, and 4 blue socks. Two socks are randomly selected without replacement. What is the probability that both socks will be white? 

This problem asks about the probability that two events will occur. Event A is selecting the first sock, while event B is selecting the second sock. The probability of both events occurring, P(A and B), is equal to P(A) times P(B). To solve this problem, start by finding P(A). There are 12 white socks, and 23 total socks. P(A) is equal to \(\frac{12}{23}\).

To find the probability of event B occurring, we need to reread the prompt. Notice that it states that the socks are selected without replacement. This means that, after the first sock is chosen, it is not placed back into the sock drawer. If the sock had been replaced, P(B) would be equal to P(A). In this case, however, P(B) will be different because there is no replacement. After one white sock is chosen, 11 white socks and 22 total socks remain in the drawer. P(B) is \(\frac{11}{22}\), or \(\frac{1}{2}\). This concept of replacement is an important one, so be sure to read the prompt carefully and think through each question.

Finally, solve for the probability of both events occurring by multiplying both fractions together.

\[P(A\space and\space B) \space=\space P(A)\space \times \space P(B)\] \[P(A\space and\space B)\space =\space \frac{12}{23}\times \frac{1}{2}\space =\space \frac{12}{46}\space = \space\frac{1}{3}\]

The probability that both socks are white is one-third, or about 33 percent. 

Another important concept related to probability has to do with how a question is worded. If a question asks for the probability that event A AND event B occur, then you multiply the probability of event A and event B. On the other hand, if a question asks for the probability of event A OR event B, then you add the probabilities. To illustrate this principle, let’s walk through an example. 

A standard deck of cards contains 52 cards. What is the probability that a randomly selected card is a queen or its suit is spades? 

This question requires you to be familiar with the makeup of a standard deck of cards, and you need to know this for the QR section. A standard card deck (without jokers) consists of 52 cards, half of which are red and the other half black. There are four suits (hearts, spades, diamonds, and clubs) with 13 cards each. Hearts and diamonds are red, while spades and clubs are black. Each suit has one card each that is 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace. Kings, queens, and jacks are referred to as face cards. 

Now, back to the question. One card is drawn, and we are tasked with finding the probability that it is a queen or a spade. There are four suits of 13 cards, each of which only contains one queen. The probability that the card is a queen will be event A, while event B is the probability that the card is a spade. P(A) is equal to \(\frac{4}{52}=\frac{1}{12}\), while P(B) equals \(\frac{13}{52}=\frac{1}{4}\). Because the question asks for the probability of event A OR event B, we need to add both probabilities.

\[\frac{1}{12}+\frac{1}{4}=\frac{4}{12}=\frac{1}{3}\]

The answer to our question is one third, or about 33%. If the question had asked for the probability that a single card is both a queen AND a spade, then the probabilities would have been multiplied, giving us a probability of \(\frac{1}{52}\). This makes sense, as a deck of cards only contains one queen of spades.

Another type of probability question you’ll likely encounter uses dice as its topic. Dice questions usually look something like the question below. 

Two fair, six-sided dice are rolled. What is the probability that both dice are a 3 or 4? 

Because there are six sides, each die has six combinations. Most questions only use two dice, both of which are six-sided. Therefore, the total number of combinations is 6 x 6 = 36 combinations. Listed below are the combinations that can sum to each total. You don’t need to have these values memorized, exactly, but you should be able to think through how many ways each number can be made. 

12 : (6, 6)

11 : (6, 5), (5, 6)

10 : (6, 4), (5, 5), (4, 6)

9 : (6, 3), (5, 4), (4, 5), (3, 6)

8 : (2, 6), (3, 5), (4, 4), (5, 3), (2, 6)

7 : (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

6 : (1, 5), (2, 4), (3, 3), (4, 2), (1, 5)

5 : (1, 4), (2, 3), (3, 2), (4, 1)

4 : (1, 3), (2, 2), (3, 1)

3 : (1, 2), (2, 1)

2 : (1, 1)

Now, back to the problem. For the example above, we need to find the probability that both dice are a 3 or 4. To do so, recognize that each die has a probability of \(\frac{2}{6}=\frac{1}{3}\) to land on a 3 or 4. The probabilities will be multiplied together because the prompt asks for the probability of both dice being 3 or 4. \(\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}\) , so one-ninth is the correct answer.

b) Combinations and permutations

Combinations and permutations are both techniques to determine the number of possible arrangements for a set of objects or numbers. The key difference between the two is ordering. If the order of objects or numbers does not matter, combinations are used. When the order does matter, however, a permutation is used. 

What exactly does ordering even mean? Imagine you have three fruits (an apple, a mango, and a banana) in front of you, and three plates in a row to place them on. You can place the fruits in any order that you choose, so you place the mango on the left, apple on the right, and banana in the middle. Your friend rearranges the fruits so that the apple is on the left plate, the banana is still in the middle, and now the mango is on the right plate. In both instances, you used the same fruits. Therefore, there was one combination of fruits; rearranging the order of the fruits doesn’t change how many combinations there are, as the fruits are the same regardless. However, because the order the fruits are lined up changes, there are two permutations. Even though the fruits are the same, they are arranged differently, leading to more than one permutation. 

How many ways can a group of items, such as the three example fruits, be ordered? To express this mathematically, we need to use a factorial operation, represented by the exclamation mark (!). Factorials are defined as:

n! = n · (n − 1) · (n − 2) · (n − 3) · ... · 2 · 1

For example, 5! is 

5! = 5 · 4 · 3 · 2 · 1 = 120 

In the previous example, there were three fruits, so there are 3! = 6 different ways to order the fruits. 

Factorials are used in the mathematical representations of combinations and permutations. Here are the formulas for each: 

Number of combinations = \(\frac{n!}{x!(n-x)!}\)
Number of permutations = \(\frac{n!}{(n-x)!}\)

n is the number of total objects
x is the number of selected objects

Let’s walk through a few examples of both combinations and permutations so you can practice using these formulas. 

A collector offers you a deal on 3 baseball cards. He has 5 cards, and he will let you choose the 3 cards you want to purchase. How many different combinations of baseball cards can you choose? 

This is a combination problem, as it doesn’t matter which order you select the three baseball cards. Therefore, we will use the combination formula. n is the number of total objects, or baseball cards in this problem (5). x is the number of objects you select, which is 3 cards. Plug these numbers into the formula and solve as follows:

Number of different combinations = \(\frac{5!}{3!(5-3)!}\)
\(=\frac{5!}{3! \times 2!}\)
\(=\frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times2 \times 1}\)
\(=\frac{120}{12}=10\)

You’ll have a calculator on the QR section, but it won’t have the factorial function. For more complicated factorials, you likely won’t have to perform any calculations. However, you do need to know how to simplify and divide factorials. Here is an example of the type of factorial you will be expected to solve.

\[\frac{8!}{6!}\]

By expanding this factorial, you can see that many numbers are found in both the numerator and the denominator. 

\[\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}\]

Because of this, the entire denominator can be canceled out, and you’re left with the equation below.

\[8 \times 7 = 56\]

Now, back to combinations and permutations. 

There are 7 runners in a certain race. Awards are given out for first, second, and third place. How many different ways can the awards be given out? 

For this problem, an award is given out based on the order that the racers finish. Order matters, so this is a permutation question. n is 7 because that is the total number of runners, and x is 3 because that is the number of awards given out. 

Number of permutations \(=\frac{7!}{(7-3)!}\)

Number of permutations \(=\frac{7!}{4!}\)

Number of permutations \(=7\times6\times5=210\)

There are 210 different ways that the awards can be given out in this race. 

Another type of combination/permutation question that frequently shows up on the DAT relates to letter arrangements. Take the problem below as an example. 

How many ways can the letters MISSISSIPPI be arranged? 

To solve letter problems like this, we need to use factorials. Start by counting the total number of letters. In this example, there are 11 letters. 11! will be the numerator of our equation. Now, count how many of each letter are repeated. MISSISSIPPI has four I’s, four S’s, and two P’s. Make each of these numbers a factorial and multiply them together as the denominator of the equation.

\[\frac{\#\space total\space letters}{(\#\space repeated \space letters)(\#\space repeated\space letters) ...}\] \[\frac{11!}{4! \times 4! \times 2!}\] \[\frac{11!}{4! \times 4! \times 2!} =34,650\space arrangements\]

An important thing to remember with letter problems like this one is that, if there are no repeated letters, the number of arrangements is equal to the factorial of the number of total letters.

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Part 3: Statistics

a) Mean, mode, and median

Questions asking about the mean, median, mode, and range should be fairly straightforward once we know the definitions. Consider the set of numbers

x1, x2, x3, ..., xn

The mean and the average can be used interchangeably. The mean is the sum of all the values divided by the number of values. Mathematically, the mean (µ) is

\[\mu =\frac{x_1+x_2+x_3+...+x_n}{n}\]

Let’s do a more concrete example. Consider the set

\[\mu =\frac{1+3+1+3+5+23+0+0+0}{9}=\frac{36}{9}=4\]

The mode is the most common value of a data set. A simple example will illustrate how straightforward calculating the mode is. The key to these questions is to keep a tally of the occurrences of each value. Let’s try this out with the same data set as before. 

1, 3, 1, 3, 5, 23, 0, 0, 0 

The first value is 1, and it occurs twice, so let’s keep track of this with 

1 : 2

The next value is 3, and it also occurs twice, so we add 3 : 2. We now have

{1 : 2}, {3 : 2} 

As we keep on going, we get 

{1 : 2}, {3 : 2}, {5 : 1}, {23 : 1}, {0 : 3}

We see that 0 has the greatest number of occurrences at 3, thus our mode is 0. For the median and range, we have to order the set of numbers, thus we will take the set of x1, x2, x3, ..., xn and turn it into

x(1), x(2), x(3), ..., x(n)

such that

x(1) ≤ x(2) ≤ x(3) ≤ ...x(n−1) ≤ x(n)

where x(i) is the ith smallest value.

Median is the middle value of the ordered list. Let us assume that n is odd, so the median is defined as

Median=\(x_{(\frac{n-1}{2}+1)}\)

Let’s do some practice using this set of data  

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