Probability: Videos & Practice Problems

Probability is a fundamental concept used to assess the likelihood of events, calculated through theoretical and empirical methods. Theoretical probability is determined before events occur, while empirical probability is based on actual outcomes. Events can be mutually exclusive, where they cannot happen simultaneously, or non-mutually exclusive, where overlap exists. For independent events, the probability of both occurring is found by multiplying their individual probabilities. Understanding these concepts is crucial for analyzing real-world scenarios, such as weather forecasts or game outcomes.

Introduction to Probability

Probability is a concept we encounter daily, whether checking the weather forecast or contemplating lottery odds. It can be calculated mathematically, allowing us to quantify the likelihood of various events. In probability notation, we denote the probability of an event as P(event). An event can be any occurrence, such as the chance of rain or flipping heads on a coin.

There are two primary types of probability: theoretical and empirical. Theoretical probability is based on possible outcomes before any events occur. For instance, when flipping a coin, the theoretical probability of landing heads is calculated as:

$$P(\text{heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{2}$$

In contrast, empirical probability is derived from actual experiments or observations. If you flip a coin three times and get heads twice, the empirical probability is:

$$P(\text{heads}) = \frac{\text{Number of heads observed}}{\text{Total flips}} = \frac{2}{3}$$

To illustrate further, consider rolling a six-sided die. The probability of rolling a number greater than 3 can be calculated as follows:

$$P(\text{number} > 3) = \frac{\text{Number of outcomes greater than 3}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} = 0.5$$

When using empirical data, if you roll the die 10 times and find that 8 rolls are greater than 3, the empirical probability becomes:

$$P(\text{number} > 3) = \frac{8}{10} = \frac{4}{5} = 0.8$$

The difference between theoretical and empirical probabilities often arises from sample size. A larger number of trials will yield results closer to the theoretical probability. In probability studies, all possible outcomes of an event can be represented as a sample space, denoted in set notation. For example, the sample space for flipping a coin is:

$$S = \{ \text{heads, tails} \}$$

Understanding these foundational concepts of probability equips you to analyze and interpret data effectively, whether in academic settings or real-world applications.

Given the data below, determine the probability that a person randomly selected from Group 1 will be wearing jeans.

• A. 0.37
• B. 0.48
• C. 0.52
• D. 0.63

Answer: (C)

In your coin purse, you have 3 quarters, 4 nickels, & 2 dimes. If you pick a coin at random, what is the probability that it will be a quarter?

• A. 0.33
• B. 0.44
• C. 0.50
• D. 0.66

Answer: (A)

Introduction to Probability Example

In a lottery game where players must select five numbers from a pool of 40 (ranging from 1 to 40), calculating the probability of winning involves understanding combinations and the total number of possible outcomes. The combinations formula, denoted as \( nCr \), is essential for determining how many ways you can choose \( r \) objects from \( n \) total objects without regard to the order of selection.

To find the total number of combinations of 5 numbers from 40, we use the formula:

\[\binom{n}{r} = \frac{n!}{(n-r)! \cdot r!}\]

In this case, \( n = 40 \) and \( r = 5 \). Plugging in these values, we calculate:

\[\binom{40}{5} = \frac{40!}{(40-5)! \cdot 5!} = \frac{40!}{35! \cdot 5!}\]

To simplify, we can express the numerator as:

\[40 \times 39 \times 38 \times 37 \times 36 \times 35!\]

After canceling \( 35! \) from the numerator and denominator, we are left with:

\[\frac{40 \times 39 \times 38 \times 37 \times 36}{5!}\]

Next, we expand \( 5! \) as \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Performing the multiplication in the numerator and dividing by \( 120 \) yields a total of:

\[658,008\]

This number represents the total possible combinations of selecting 5 numbers from 40.

To calculate the probability of winning with one lottery ticket, we take the number of successful outcomes (1 winning combination) and divide it by the total combinations:

\[P(\text{winning with 1 ticket}) = \frac{1}{658,008} \approx 0.000152\]

This indicates a very low probability of winning with a single ticket.

If a player purchases 50 different lottery tickets, the probability of winning increases. The calculation for this scenario is:

\[P(\text{winning with 50 tickets}) = \frac{50}{658,008} \approx 0.00076\]

While this is a higher probability than purchasing just one ticket, it still reflects the challenging odds of winning the lottery. Understanding these calculations helps players grasp the likelihood of winning in such games.

Complementary Events

Understanding probability involves not only calculating the likelihood of an event occurring but also determining the probability that an event will not happen. This concept is encapsulated in the idea of the complement of an event. For instance, when rolling a six-sided die, if we define the event of rolling a 4 as event A, the complement of A includes all outcomes where a 4 is not rolled, specifically rolling a 1, 2, 3, 5, or 6.

The notation for the complement of an event can vary; it may be represented as \( A' \), \( \overline{A} \), or \( \neg A \). To calculate the probability of event A occurring, we use the formula:

\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} \]

In the case of rolling a die, the probability of rolling a 4 is:

\[ P(A) = \frac{1}{6} \]

To find the probability of not rolling a 4, we can count the outcomes that do not include a 4, which are 1, 2, 3, 5, and 6. Thus, the probability of the complement of A is:

\[ P(A') = \frac{5}{6} \]

It is essential to note that the sum of the probabilities of an event and its complement always equals 1:

\[ P(A) + P(A') = 1 \]

This relationship leads us to a useful formula for calculating the probability of an event not occurring:

\[ P(A') = 1 - P(A) \]

For example, when drawing a card from a standard deck of 52 cards, if we want to find the probability of not drawing a queen, we first calculate the probability of drawing a queen. There are 4 queens in the deck, so:

\[ P(\text{Queen}) = \frac{4}{52} \]

Using the complement formula, the probability of not drawing a queen becomes:

\[ P(\text{Not Queen}) = 1 - P(\text{Queen}) = 1 - \frac{4}{52} = \frac{48}{52} \approx 0.92 \]

This method simplifies the process of finding probabilities by allowing us to leverage the known probability of an event to easily determine its complement. Practicing these calculations will enhance your understanding of probability and its applications.

When drawing a marble out of a bag of red, green, and yellow marbles 8 times, a red or yellow marble is drawn 6 times. What is the probability of drawing a green marble?

• A. 0.025
• B. 0.125
• C. 0.25
• D. 0.75

Answer: (C)

A weatherman states that the probability that it will rain tomorrow is 10%, or 0.1, & the probability that it will snow is 25%, or 0.25. What is the probability that it will not rain or snow?

• A. 0.35
• B. 0.65
• C. 0.75
• D. 0.90

Answer: (B)

Probability of Mutually Exclusive Events

Understanding the probability of multiple events is essential in probability theory, especially when distinguishing between events that can occur simultaneously and those that cannot. Events that cannot happen at the same time are termed mutually exclusive. For instance, if you consider the events of wearing a blue shirt (Event A) and wearing a green shirt (Event B), these events are represented by separate circles in a Venn diagram, indicating no overlap. This means you can either wear a blue shirt or a green shirt, but not both at the same time.

In contrast, events that can occur together are not mutually exclusive. For example, wearing a blue shirt (Event A) and wearing green pants (Event B) can happen simultaneously, as illustrated by overlapping circles in a Venn diagram. This overlap signifies that both events can occur at the same time.

To further clarify, consider flipping a coin. The outcomes of getting heads or tails are mutually exclusive since both cannot occur in a single flip. Conversely, when rolling a die, the event of rolling a 6 and the event of rolling a number higher than 3 are not mutually exclusive, as rolling a 6 satisfies both conditions.

When calculating the probability of mutually exclusive events, the approach is straightforward: you simply add the probabilities of each event. This is often referred to as or probability, denoted by the symbol ∪ in set notation. For example, if you want to find the probability of rolling a 3 or a 5 on a six-sided die, you would calculate it as follows:

Let \( P(A) \) be the probability of rolling a 3, and \( P(B) \) be the probability of rolling a 5. Since there is one way to roll a 3 and one way to roll a 5, the probabilities are:

\( P(A) = \frac{1}{6} \) and \( P(B) = \frac{1}{6} \)

Adding these probabilities gives:

\( P(A \cup B) = P(A) + P(B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \)

This result indicates that the probability of rolling either a 3 or a 5 is \( \frac{1}{3} \), which is approximately 0.33. Understanding these concepts allows for a clearer grasp of how to approach problems involving multiple events in probability.

If a single card is randomly selected from a deck of cards, what is the probability of selecting an ace or a king?

• A. 0.0059
• B. 0.077
• C. 0.15
• D. 0.85

Answer: (C)

For two mutually exclusive events A and B, compute $P\left(A\cup B\right)$ if $P\left(A\right)=0.15$ and $P\left(B\right)=0.32$

• A. 0.048
• B. 0.17
• C. 0.47
• D. 0.53

Answer: (C)

Probability of Non-Mutually Exclusive Events

Understanding the calculation of the probability of two events occurring is essential in probability theory, especially when distinguishing between mutually exclusive and non-mutually exclusive events. For mutually exclusive events, the probability of either event A or event B occurring is straightforward: you simply add their individual probabilities together. However, when dealing with non-mutually exclusive events, an additional step is required due to the overlap where both events can occur simultaneously.

To calculate the probability of event A or event B for non-mutually exclusive events, you start by adding the probabilities of each event. However, since the overlap (where both events occur) has been counted twice, you must subtract the probability of the overlap. This can be expressed with the formula:

$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$

In this formula, \( P(A \cup B) \) represents the probability of either event A or event B occurring, \( P(A) \) is the probability of event A, \( P(B) \) is the probability of event B, and \( P(A \cap B) \) is the probability of both events occurring simultaneously.

To illustrate this, consider the example of rolling a six-sided die. Let’s define two events: event A is rolling a number greater than 3, and event B is rolling an even number. The possible outcomes when rolling the die are 1, 2, 3, 4, 5, and 6. Analyzing these outcomes:

• Event A (greater than 3): Possible outcomes are 4, 5, and 6 (3 outcomes).
• Event B (even numbers): Possible outcomes are 2, 4, and 6 (3 outcomes).
• Overlap (both greater than 3 and even): Possible outcomes are 4 and 6 (2 outcomes).

Now, applying the formula:

1. Calculate \( P(A) = \frac{3}{6} \) (for outcomes 4, 5, 6).

2. Calculate \( P(B) = \frac{3}{6} \) (for outcomes 2, 4, 6).

3. Calculate \( P(A \cap B) = \frac{2}{6} \) (for outcomes 4, 6).

Substituting these values into the formula gives:

$$ P(A \cup B) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3} $$

As a decimal, this probability is approximately 0.67. This result indicates that there is a 67% chance of rolling a number that is either greater than 3 or an even number.

In summary, the general formula for calculating the probability of two events, whether they are mutually exclusive or not, remains consistent. For mutually exclusive events, the overlap term is zero, simplifying the calculation. This understanding allows for a comprehensive approach to probability, enabling the calculation of outcomes in various scenarios.

Probability of Non-Mutually Exclusive Events Example

In this scenario, we are tasked with determining the probability that a randomly selected individual from a group of 300 people is either wearing shorts or a green shirt. To approach this, we first identify the relevant data: 188 individuals are wearing shorts, and 106 are wearing a green shirt. However, since some individuals may be wearing both, we need to account for this overlap to avoid double counting.

We define two events: let event A represent wearing shorts and event B represent wearing a green shirt. The formula for calculating the probability of either event A or event B occurring, when the events are not mutually exclusive, is given by:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Here, P(A ∪ B) is the probability of either event A or event B occurring, P(A) is the probability of wearing shorts, P(B) is the probability of wearing a green shirt, and P(A ∩ B) is the probability of wearing both.

Using the data provided, we can calculate each probability:

• P(A) = 188/300
• P(B) = 106/300
• P(A ∩ B) = 89/300

Substituting these values into the formula gives us:

P(A ∪ B) = (188/300) + (106/300) - (89/300)

Calculating this step-by-step:

P(A ∪ B) = (188 + 106 - 89) / 300 = 205 / 300

To simplify this fraction, we divide both the numerator and the denominator by 5, resulting in:

P(A ∪ B) = 41 / 60

For those who prefer a decimal representation, this probability can also be expressed as approximately 0.68. Therefore, the probability that a randomly selected person from this group is wearing either shorts or a green shirt, or both, is 0.68.

A card is drawn from a standard deck of 52 cards. What is the probability that the card is a diamond or a king?

• A. 0.33
• B. 0.31
• C. 0.15
• D. 0.85

Answer: (B)

Probability of Multiple Independent Events

Understanding the concept of probability is essential in statistics, particularly when dealing with independent events. When calculating the probability of two events occurring together, such as flipping a coin twice and getting heads both times, we refer to this as "and probability." This type of probability is calculated when the outcome of one event does not influence the outcome of another, which is the hallmark of independent events.

Two events are considered independent if the occurrence of one does not affect the occurrence of the other. For example, flipping a coin twice results in independent events because the outcome of the first flip (heads or tails) does not impact the outcome of the second flip. Conversely, if you draw a marble from a bag and keep it, the probability of drawing a second marble is affected by the first draw, making these events dependent.

To calculate the probability of two independent events A and B occurring together, you multiply their individual probabilities. This is expressed mathematically as:

P(A \text{ and } B) = P(A) \times P(B)

For instance, when calculating the probability of getting heads on two consecutive coin flips, the probability of getting heads on the first flip is \( \frac{1}{2} \) and the same for the second flip. Thus, the combined probability is:

P(\text{Heads on 1st flip and Heads on 2nd flip}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \text{ or } 0.25.

Similarly, when rolling a six-sided die, the probability of rolling an even number (2, 4, or 6) is \( \frac{3}{6} \) or \( \frac{1}{2} \), and the probability of rolling a 3 is \( \frac{1}{6} \). Therefore, the probability of rolling an even number followed by a 3 is:

P(\text{Even number and 3}) = \frac{3}{6} \times \frac{1}{6} = \frac{3}{36} = 0.08.

For scenarios involving more than two independent events, the same principle applies: simply multiply the probabilities of all events together. This foundational understanding of independent events and their probabilities is crucial for further studies in statistics and probability theory.

The spinner below has 6 equal regions. Find the probability of landing on yellow for the first spin and not landing on yellow on the second spin.

• A. 0.11
• B. 0.22
• C. 0.66
• D. 0.88

Answer: (B)

The spinner below has 6 equal colored regions numbered 1-6. Find the probability of stopping on yellow for the first spin, stopping on an even number on the second spin, and stopping on blue or red on the third spin.

• A. 0.11
• B. 0.17
• C. 0.50
• D. 0.89

Answer: (A)

Here’s what students ask on this topic:

What is the difference between theoretical and empirical probability?

Theoretical probability is calculated based on the possible outcomes of an event before it actually happens. It is determined by dividing the number of favorable outcomes by the total number of possible outcomes. For example, the theoretical probability of getting heads in a coin flip is 1/2. Empirical probability, on the other hand, is based on actual data from experiments or observations. It is calculated by dividing the number of times an event occurs by the total number of trials. For instance, if you flip a coin 10 times and get heads 7 times, the empirical probability of getting heads is 7/10. The key difference is that theoretical probability is based on expected outcomes, while empirical probability is based on observed outcomes.

How do you calculate the probability of an event not happening?

The probability of an event not happening, also known as the complement of the event, can be calculated using the formula:

$P(A\prime )=1-P\left(A\right)$

Here, $P\left(A\right)$ is the probability of the event happening, and $P(A\prime )$ is the probability of the event not happening. For example, if the probability of drawing a queen from a deck of 52 cards is 4/52, then the probability of not drawing a queen is:

$1-\frac{4}{52}=\frac{48}{52}=0.92$

What are mutually exclusive events in probability?

Mutually exclusive events are events that cannot happen at the same time. If one event occurs, the other cannot. For example, when flipping a coin, getting heads and getting tails are mutually exclusive events because you cannot get both heads and tails on a single flip. In a Venn diagram, mutually exclusive events are represented by non-overlapping circles. The probability of either event A or event B occurring (but not both) is calculated by adding their individual probabilities:

$P(A\cup B)=P\left(A\right)+P\left(B\right)$

For example, if the probability of rolling a 3 on a die is 1/6 and the probability of rolling a 5 is also 1/6, then the probability of rolling a 3 or a 5 is:

$\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$

How do you calculate the probability of independent events?

To calculate the probability of two independent events both occurring, you multiply their individual probabilities. Independent events are those where the outcome of one event does not affect the outcome of the other. The formula is:

$P(A\cap B)=P\left(A\right)\times P\left(B\right)$

For example, if the probability of getting heads on a coin flip is 1/2 and the probability of rolling a 3 on a six-sided die is 1/6, then the probability of both getting heads and rolling a 3 is:

$\frac{1}{2}\times \frac{1}{6}=\frac{1}{12}=0.083$

This means there is an 8.3% chance of both events occurring.

What is the sample space in probability?

The sample space in probability is the set of all possible outcomes of an experiment. It is often denoted by the letter S and is represented using set notation with curly brackets. For example, the sample space for flipping a coin is:

$S=\{\mathrm{Heads},\mathrm{Tails}\}$

Similarly, the sample space for rolling a six-sided die is:

$S=\{1,2,3,4,5,6\}$

Understanding the sample space is crucial for calculating probabilities, as it helps identify all possible outcomes and ensures that the probabilities of all outcomes sum to 1.