7.4 Addition Theorem of Probability

The Addition Theorem of Probability is a fundamental concept used to calculate the probability that at least one of several events occurs. In simpler terms, it helps us find the probability of the union of two or more events.

A crucial aspect before applying this theorem is to determine whether the events are mutually exclusive or overlapping, as the formula varies based on this distinction.

What is the Addition Theorem?

Let $E_1$ and $E_2$ be two events from a sample space. The Addition Theorem allows us to calculate the probability that either event $E_1$ or event $E_2$ (or both) happens, denoted as $P(E_1 \cup E_2)$ or $P(E_1 \text{ or } E_2)$.

1. Addition Theorem for Mutually Exclusive Events

Definition: Events are considered mutually exclusive if they cannot occur simultaneously. If one event happens, the other event cannot happen.

Formula: For two mutually exclusive events, the probability that either $E_1$ or $E_2$ occurs is the sum of their individual probabilities:

$$P(E_1 \text{ or } E_2) = P(E_1) + P(E_2)$$

For more than two mutually exclusive events: If you have multiple mutually exclusive events ($E_1, E_2, \dots, E_n$), the formula extends as follows:

$$P(E_1 \text{ or } E_2 \text{ or } \dots \text{ or } E_n) = P(E_1) + P(E_2) + \dots + P(E_n)$$

Collectively Exhaustive Mutually Exclusive Events: If a set of mutually exclusive events are also collectively exhaustive (meaning they cover the entire sample space and one of them is guaranteed to occur), their total probability is 1:

$$P(E_1 \text{ or } E_2 \text{ or } \dots \text{ or } E_n) = 1$$

Example: Consider rolling a standard six-sided die once. Let Event A be "getting a 4". Let Event B be "getting a 5".

Since a single die roll cannot result in both a 4 and a 5 at the same time, events A and B are mutually exclusive.

• $P(A) = \frac{1}{6}$ (The probability of rolling a 4)
• $P(B) = \frac{1}{6}$ (The probability of rolling a 5)

Using the Addition Theorem for mutually exclusive events:

$P(A \text{ or } B) = P(A) + P(B)$ $P(A \text{ or } B) = \frac{1}{6} + \frac{1}{6}$ $P(A \text{ or } B) = \frac{2}{6}$ $P(A \text{ or } B) = \frac{1}{3}$

Therefore, the probability of getting either a 4 or a 5 on a single die roll is $\frac{1}{3}$.

2. Addition Theorem for Overlapping Events

Definition: Events are considered overlapping (or non-mutually exclusive) if they can occur simultaneously. When calculating the probability of either event occurring, we must subtract the probability of both events happening to avoid double-counting the common outcome(s).

Formula: For two overlapping events, the probability that either $E_1$ or $E_2$ occurs is:

$$P(E_1 \text{ or } E_2) = P(E_1) + P(E_2) - P(E_1 \text{ and } E_2)$$

Here, $P(E_1 \text{ and } E_2)$ represents the probability of the intersection of $E_1$ and $E_2$, denoted as $P(E_1 \cap E_2)$.

For three overlapping events: For three overlapping events ($E_1, E_2, E_3$), the formula becomes:

$$P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(E_2) + P(E_3) - P(E_1 \cap E_2) - P(E_1 \cap E_3) - P(E_2 \cap E_3) + P(E_1 \cap E_2 \cap E_3)$$

Example: Consider a group of 50 students.

• 20 students like Football.
• 30 students like Cricket.
• 10 students like both Football and Cricket.

Let Event F be "a student likes Football". Let Event C be "a student likes Cricket".

We can calculate the probabilities:

• $P(F) = \frac{20}{50}$ (The probability a student likes Football)
• $P(C) = \frac{30}{50}$ (The probability a student likes Cricket)
• $P(F \text{ and } C) = \frac{10}{50}$ (The probability a student likes both Football and Cricket)

Applying the Addition Theorem for overlapping events:

$P(F \text{ or } C) = P(F) + P(C) - P(F \text{ and } C)$ $P(F \text{ or } C) = \frac{20}{50} + \frac{30}{50} - \frac{10}{50}$ $P(F \text{ or } C) = \frac{50}{50} - \frac{10}{50}$ $P(F \text{ or } C) = \frac{40}{50}$ $P(F \text{ or } C) = \frac{4}{5}$

Therefore, the probability that a randomly selected student likes either Football or Cricket is $\frac{4}{5}$.

Summary of Formulas

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Interview Questions Related to the Addition Theorem

1. What is the Addition Theorem of Probability? The Addition Theorem of Probability is used to find the probability that at least one of several events occurs, essentially calculating the probability of the union of events.

2. How does the addition rule differ for mutually exclusive and overlapping events? For mutually exclusive events, the probability of their union is simply the sum of their individual probabilities ($P(A \cup B) = P(A) + P(B)$). For overlapping events, the probability of their union is the sum of their individual probabilities minus the probability of their intersection ($P(A \cup B) = P(A) + P(B) - P(A \cap B)$) to avoid double-counting.

3. Define mutually exclusive events with an example. Mutually exclusive events are events that cannot happen at the same time. For example, when rolling a single die, rolling a 3 and rolling a 5 are mutually exclusive events.

4. Define overlapping events with an example. Overlapping events are events that can happen at the same time. For example, drawing a card from a standard deck, drawing a King and drawing a Heart are overlapping events, as the King of Hearts is common to both.

5. What does it mean for events to be collectively exhaustive? Events are collectively exhaustive if they cover all possible outcomes in the sample space, meaning that at least one of these events is guaranteed to occur.

6. State the addition theorem formula for two overlapping events. For two overlapping events $E_1$ and $E_2$, the formula is $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$.

7. Give the formula for three overlapping events in probability. For three overlapping events $E_1, E_2, E_3$, the formula is $P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(E_2) + P(E_3) - P(E_1 \cap E_2) - P(E_1 \cap E_3) - P(E_2 \cap E_3) + P(E_1 \cap E_2 \cap E_3)$.

8. When can we say that $P(E_1 \cup E_2) = P(E_1) + P(E_2)$? We can say that $P(E_1 \cup E_2) = P(E_1) + P(E_2)$ when events $E_1$ and $E_2$ are mutually exclusive, meaning they cannot occur at the same time ($P(E_1 \cap E_2) = 0$).