Combining Events Using OR

In probability and logic theory, combining events using OR (denoted as $\cup$) means considering situations where either one event occurs, or the other occurs, or both occur. This is known as the “inclusive OR.”

Given two events $A$ and $B$, one can define the event $A$ or $B$ to be the event that at least one of the events $A$ or $B$ occurs. The probability of the events $A$ or $B$ using the Addition Rule of probability can be computed easily. Learn the Basics of Probability.

Addition Rule of Probability (for Non-Mutually Exclusive Events)

If $A$ and $B$ are two events for an experiment, then
$$P(A\,\, or \,\,B) = P(A\cup B) = P(A) + P(B) – P(A\,\,and \,\, B)$$
This accounts for the overlap between events to avoid double-counting

Addition Rule Probability (for Mutually Exclusive Events)

Two events are called mutually exclusive events if both events cannot occur at the same time (cannot occur together). In this case, when the mutually exclusive events, $P(A\,\,\cap\,\,B)=0$, so the addition rule simplies to:
$$P(A\,\,or\,\,B) = P(A) + P(B)$$
This does not account for the overlap between events to avoid double-counting.

Combining Events using OR

Real Life Examples of Combining Events using OR

The following are a few real-life examples of Combining Events Using OR.

Weather Forecast Example

Suppose Event $A$ represents that it will rain tomorrow and Event $B$ that it will snow tomorrow. One can compute the probability that it will rain OR snow tomorrow. This means that at least one of them happens (it could be rain, snow, or both).
Suppose that the chance of rain tomorrow = $P(A)$ = 30% = 0.3. Supose that the probability of snow tomorrow = $P(B)$ = 20% = 0.2. Suppose the chances of both rain and snow are $P(A \cap B)$ = 5% = 0.5.
Therefore,
\begin{align*}
P(A \cup B) &= P(A) + P(B) – P(A \cup B) \\
& = 0.3 + 0.2 – 0.05 = 0.45
\end{align*}
There is a 45% chance that it will rain or snow tomorrow.

Job Requirements

Suppose Event $A$ represents that applicants must have a Bachelor’s degree, and Event $B$ represents that applicants must have 3 years of experience. One can compute the probability (or count) that the applicant must have a bachelor’s degree OR 3 years of experience to apply. The applicant will qualify if he/she have either one or both experiences.
Suppose there are 100 applicants for a certain job. For Event $A$, there are 40 applicants who have a Bachelor’s degree, and Event $B$ represents that there are 30 applicants who have more than 5 years of experience. Similarly, 10 applicants have both a Bachelor’s degree and have more than 3 years of experience. The number of qualifying applicants will be

\begin{align*}
A \cup B &= A + B – A \cap B \\
& = 40 + 30 – 10 = 60
\end{align*}
Therefore, 60 applicants meet at least one requirement (degree OR experience).

Restaurant Menu Choices

Consider Event $A$ represents the meal comes with fries, and Event $B$ represents the meal comes with a salad. One can compute if a customer can pick one, or sometimes both, if allowed. For illustrative purposes, suppose a Fast-Food Chain tracks 1000 orders. The Event $A$ represents 400 customers who choose fries, and Event $B$ represents 300 customers who choose a salad. Similarly, there are 100 customers who both choose fries and salad. The number of customers’ choices for both fries and salad will be

\begin{align*}
A \cup B &= A + B – A\cap B\\
&= 400 + 300 – 100 = 600
\end{align*}
600 customers ordered fries OR salad (or both).

Discount Offers

Let Event $A$ represent the use of a promo code for 10% off, Event $B$ represents a Student ID for 15% off. One uses a promo code or a Student ID to get a discount. Suppose a store offers two discount options to 200 customers. Event $A$ represents 65% of customers who used a coupon, Event $B$ represents that 13% customers showed their Student ID. 7% customers have used both the coupon and the Student ID. The probability that at least one discount is used will be

\begin{align*}
P(A \cup B) &= P(A) + P(B) – P(A \cap B)\\
& = 0.65 + 0.13 – 0.07 = 0.71
\end{align*}
71% of the customers have used at least one discount.

Security System Access

Suppose a building logs 500 entry attempts. Out of 500, 300 entries used a keycard, 200 used a PIN code, and 50 used both methods. What is the probability that both entry attempts are made?
\begin{align*}
P(A\cap B) &= P(A) + P(B) – P(A \cap B)\\
& = \frac{300}{500} + \frac{200}{500} – \frac{50}{500} = 0.6 + 0.4 – 0.1 = 0.9
\end{align*}
There are 90% ($500\times 0.9=450$) entries that used a keycard OR a PIN.

General Knowledge Quizzes

FAQs about Combining Events

  • What is meant by Combining Events?
  • What symbol is used to combine two or more events?
  • What rule of probability is used to combine events?
  • Give some real-life examples of Combining Events using OR.
  • What are mutually and Non-Mutually Exclusive Events?

MCQs Permutations Combinations 14

Test your knowledge of permutations and combinations with this interactive quiz! The MCQs Permutations Combinations Quiz covers essential concepts like factorials, combinations, arrangements, and real-world applications. This quiz is perfect for students and enthusiasts looking to sharpen their probability and counting skills. Let us start with the MCQs Permutation Combinations Quiz now.

Learn about Counting Techniques in Probability

Online MCQs Permutations Combinations with Answers

Online Quiz about Permutations and Combinations with Answers

1. How many combinations of size 4 can be formed from a set of 6 distinct objects?

 
 
 
 

2. ${}^nP_r$ =

 
 
 
 

3. The number of 3-digit telephone area codes that can be made if repetitions are not allowed is

 
 
 
 

4. A homeowner doing some remodeling requires the services of both a plumbing contractor and an electrical contractor. If there are 15 plumbing contractors and 10 electrical contractors available in the area, in how many ways can the contractors be chosen?

 
 
 
 

5. An arrangement of all or some of a set of objects in a definite order is called

 
 
 
 

6. ${}^nC_r$ =

 
 
 
 

7. Seventeen teams can take part in the Football Championship of a country. In how many ways can the Gold, Silver, and Bronze medals be distributed among the teams?

 
 
 
 

8. The $0!$ is

 
 
 
 

9. Which of the following statements is true?

 
 
 
 

10. ${}^5C_5$ is equal to

 
 
 
 

11. In how many ways can be letters in the word UNIVERSITY be arranged randomly

 
 
 
 

12. How many permutations of size 3 can be constructed from the set (A, B, C, D, E)?

 
 
 
 

13. The number of ways to select 2 persons from 6, ignoring the order of selection, is

 
 
 
 

14. ${}^{10}C_5=$

 
 
 
 

15. An experiment consists of three stages. There are five ways to accomplish the first stage, four ways to accomplish the second stage, and three ways to accomplish the third stage. The total number of ways to accomplish the experiment is

 
 
 
 

16. An arrangement of objects without caring for the order is called

 
 
 
 

17. In how many ways can a team of 6 players be chosen from 11 persons

 
 
 
 

18. The difference between permutation and combination lies in the fact that

 
 
 
 

19. How many terms are in the expansion of the $(q+p)^n$

 
 
 
 

20. $n!=$?

 
 
 
 

Question 1 of 20

Online MCQs Permutations Combinations

  • The number of ways to select 2 persons from 6, ignoring the order of selection, is
  • $n!=$?
  • An arrangement of all or some of a set of objects in a definite order is called
  • An arrangement of objects without caring for the order is called
  • ${}^nP_r$ =
  • ${}^nC_r$ =
  • In how many ways can a team of 6 players be chosen from 11 persons
  • How many terms are in the expansion of the $(q+p)^n$
  • ${}^{10}C_5=$
  • ${}^5C_5$ is equal to
  • The difference between permutation and combination lies in the fact that
  • Which of the following statements is true?
  • A homeowner doing some remodeling requires the services of both a plumbing contractor and an electrical contractor. If there are 15 plumbing contractors and 10 electrical contractors available in the area, in how many ways can the contractors be chosen?
  • How many permutations of size 3 can be constructed from the set (A, B, C, D, E)?
  • How many combinations of size 4 can be formed from a set of 6 distinct objects?
  • An experiment consists of three stages. There are five ways to accomplish the first stage, four ways to accomplish the second stage, and three ways to accomplish the third stage. The total number of ways to accomplish the experiment is
  • The $0!$ is
  • In how many ways can be letters in the word UNIVERSITY be arranged randomly
  • Seventeen teams can take part in the Football Championship of a country. In how many ways can the Gold, Silver, and Bronze medals be distributed among the teams?
  • The number of 3-digit telephone area codes that can be made if repetitions are not allowed is

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Complement of an Event

Probability is a fundamental concept in statistics used to quantify uncertainty. One of the key concepts in probability is the Complement of an event. The complement of an event provides a different perspective on computing the probabilities, that is, it is used to determine the likelihood of an event not occurring. Let us explore how the complement of an event is used for the computation of probability.

What is the Complement of an Event?

The complement of an event $E$ is denoted by $E’$, encompasses all outcomes in the sample space that are not part of event $E$. In simple terms, if event $E$ represents a specific outcome or set of outcomes, its complement represents everything else that could occur.

For example, let the event $E$ be rolling a 4 on a six-sided die; the complement of event $E$ is ($E’$) rolling a 1, 2, 3, 5, or 6.

Note that event $E$ and its complement $E’$ cover the entire sample space of the die roll.

Complement Rule: Calculating Probabilities

A pivotal property of complementary events is that the sum of their probabilities is 1 (or 100%). This is because either the event happens or it does not happen, as there are no other probabilities. It can be described as
$$P(E) + P(E’) = 1$$
This leads to the complement rule, which states that
$$P(E’)= 1- P(E)$$
It is useful when computing the probability of an event not occurring.

Complement of an Event in Probability

Examples (Finding the Complement of an Event)

Suppose the probability that today is a rainy day is 0.3. The probability of it not raining today is $$1-0.3 = 0.7$$

Similarly, the probability of rolling a 2 on a fair die is $P(E) = \frac{1}{6}$. the probability of not rolling a 2 is $P(E’)=1-\frac{1}{6} = \frac{5}{6}$.

Why use the Complement Rule?

Sometimes, calculating the probability of the complement is easier than calculating the probability of the event itself. For example,

Question: What is the probability of getting at least one head in three coin tosses?
Solution: Instead of listing all possible favourable outcomes, one can easily use the complement rule. That is,
Complement Event: Getting no heads (all tails)
Probability of all tails = $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$. Therefore, the probability of at least one head is

P(At least one head) = $1 – \frac{1}{8} = \frac{7}{8}$
This approach is quicker than counting all possible cases; that is, one can avoid enumerating all the favourable outcomes.

Properties of Complementary Events

  • Mutually Exclusive: An event and its complement cannot occur together (simultaneously)
  • Collectively Exhaustive: An event and its complement encompass all possible outcomes
  • Probability Sum: The probabilities of an event and its complement add up to 1.

Understanding complements in probability can make complex problems much simpler and easier.

Practical Applications

Understanding complements is invaluable in various fields:

  • Quality Control: Determining the probability of defects in manufacturing
  • Everyday Decisions: Estimating probabilities in daily life, such as the chance of missing a bus or the likelihood of rain.
  • Game Theory: Calculating chances of winning or losing scenarios
  • Risk Assessment: Evaluating the likelihood of adverse events not occurring

More Examples (Complement of an Event)

  • In a standard 52-card deck, what is the probability of not drawing a heart card?
    $P(Not\,\,Heart) = 1 – P(Heart) = 1 – \frac{13}{52} = \frac{39}{52}$
  • If the probability of passing an examination is 0.85, what is the probability of failing it?
    $P(Fail) = 1 – P(Pass) = 1 – 0.85 = 0.15$
  • If the probability that a flight will be delayed is 0.13, then the probability that it will not be delayed will be $1 – 0.13 = 0.87$
  • If $k$ is the event of drawing a king card from a well-shuffled 52-card deck, then the event $K’$ is the event that a king is not drawn, so $K’$ will contain 48 possible outcomes.

Data Analysis in the R Programming Language