Probability of the Union of Two Events - Definition, Examples, and Formula

Formula

The probability of the union of two events, denoted by \(P(A \cup B)\) , is the probability that at least one of the two events occurs. It is calculated as:

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

where \(P(A)\) is the probability of event \(A\) occurring, \(P(B)\) is the probability of event \(B\) occurring, and \(P(A \cap B)\) is the probability of both events \(A\) and \(B\) occurring together.

The formula works because if we add the probabilities of both events \(A\) and \(B\) , we will have counted the intersection \( A \cap B\) twice, so we need to subtract it once to get the correct probability of the union.

If two events A and B are mutually exclusive, meaning they cannot occur at the same time, then the probability of the union of A and B is simply the sum of the individual probabilities:

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

In general, two events are mutually exclusive if:

$$P(A \cap B) = 0$$

That is, if the probability of both events occurring at the same time is zero.

Example 1: (Not Mutually Exclusive Events)

Suppose we have two events \(A\) and \(B\) such that \(P(A) = 0.6\) , \(P(B) = 0.4\) , and \(P(A \cap B) = 0.2\) . To find the probability of the union of events \(A\) and \(B\) (i.e., the probability that at least one of the events occurs), we can use the formula:

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

Substituting the given values, we get:

$$P(A \cup B) = 0.6 + 0.4 - 0.2 = 0.8$$

Therefore, the probability that at least one of the events \( A\) or \(B\) occurs is \( 0.8\) , or 80%.

Example 2: (Not Mutually Exclusive Events)

Here's an example of the probability of the union of two events using LaTeX syntax:

Suppose we roll a fair six-sided die. Let event A be the roll of an even number, and event B be the roll of a number less than 4. Then the probability of the union of events A and B (i.e., rolling an even number or a number less than 4, or both) is:

$$P(A \cup B) = P(\text{even number}) + P(\text{number less than 4}) - P(\text{even number and number less than 4})$$

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

Since there are three even numbers and three numbers less than 4, but only one number (2) that satisfies both conditions, we have:

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

$$P(A \cup B) = \frac{3}{6} + \frac{3}{6} - \frac{1}{6} = \frac{5}{6}$$

Therefore, the probability of rolling an even number or a number less than 4 (or both) is 5/6.

Example 3: (Mutually Exclusive Events)

The probability of the union of two mutually exclusive events is simply the sum of their individual probabilities. In other words, if two events are mutually exclusive, it means that they cannot occur at the same time, and so the probability of their union is the sum of their probabilities.

For example, suppose you are rolling a single fair six-sided die. Event A is defined as rolling an even number (2, 4, or 6), and event B is defined as rolling an odd number (1, 3, or 5). These two events are mutually exclusive, since you cannot roll an even and an odd number at the same time. Therefore, the probability of rolling either an even or an odd number is simply the sum of the probabilities of rolling each separately:

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

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

Note that the sum of the probabilities in this case is equal to 1, since rolling an even or an odd number are the only possible outcomes.