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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.
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).
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?
