Understanding Probability with Mutually Exclusive Events

The question asks us to find the conditional probability \(P(A | (A \cup B))\) for two mutually exclusive events A and B, given \(P(A) = 0.2\) and \(P(A̅ \cap B) = 0.3\).

Let's break down the information given and the terms used:

• Mutually Exclusive Events: Two events A and B are mutually exclusive if they cannot occur at the same time. This means their intersection is empty, i.e., \(A \cap B = \emptyset\). Consequently, the probability of their intersection is zero: \(P(A \cap B) = 0\).
• Given Probabilities:
  • \(P(A) = 0.2\)
  • \(P(A̅ \cap B) = 0.3\). This represents the probability that event B occurs and event A does not occur.

Analyzing \(P(A̅ \cap B)\) for Mutually Exclusive Events

Since A and B are mutually exclusive, if event B occurs, event A cannot occur. This means that the set of outcomes where B happens and A does not happen (\(A̅ \cap B\)) is simply the set of outcomes where B happens, because any outcome in B is automatically not in A. Therefore, \(A̅ \cap B = B\).

So, the given probability \(P(A̅ \cap B) = 0.3\) implies \(P(B) = 0.3\).

Let's summarize the probabilities we have:

Calculating \(P(A \cup B)\)

For mutually exclusive events A and B, the probability of their union (A or B occurring) is the sum of their individual probabilities:

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

Substituting the values we have:

\(P(A \cup B) = 0.2 + 0.3 = 0.5\)

Calculating the Conditional Probability \(P(A | (A \cup B))\)

The formula for conditional probability \(P(X | Y)\) is \(\frac{P(X \cap Y)}{P(Y)}\). In this question, \(X = A\) and \(Y = A \cup B\). So, we need to calculate:

\(P(A | (A \cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}\)

Let's simplify the term \(A \cap (A \cup B)\). The intersection of event A with the union of A and B contains all outcomes that are in A AND are also in either A or B. The outcomes that satisfy this condition are simply the outcomes that are in A. Therefore, \(A \cap (A \cup B) = A\).

Now, substitute this back into the conditional probability formula:

\(P(A | (A \cup B)) = \frac{P(A)}{P(A \cup B)}\)

We have \(P(A) = 0.2\) and we calculated \(P(A \cup B) = 0.5\).

Substituting these values:

\(P(A | (A \cup B)) = \frac{0.2}{0.5} = \frac{\frac{2}{10}}{\frac{5}{10}} = \frac{2}{5}\)

Thus, the conditional probability \(P(A | (A \cup B))\) is \(\frac{2}{5}\).

Step-by-Step Solution Summary

1. Identify that A and B are mutually exclusive, so \(P(A \cap B) = 0\).
2. Use \(P(A̅ \cap B) = 0.3\) to find \(P(B)\). Since A and B are mutually exclusive, \(A̅ \cap B = B\), so \(P(B) = 0.3\).
3. Calculate \(P(A \cup B)\) using the formula for mutually exclusive events: \(P(A \cup B) = P(A) + P(B) = 0.2 + 0.3 = 0.5\).
4. Simplify the intersection term in the conditional probability formula: \(A \cap (A \cup B) = A\).
5. Apply the conditional probability formula: \(P(A | (A \cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)}\).
6. Substitute the known values: \(P(A | (A \cup B)) = \frac{0.2}{0.5} = \frac{2}{5}\).

Revision Table: Probability Concepts

Additional Information: Set Operations in Probability

In probability, events are treated as sets of outcomes. Understanding basic set operations helps in solving problems.

• Union (\(\cup\)): \(A \cup B\) is the event where A occurs, or B occurs, or both occur.
• Intersection (\(\cap\)): \(A \cap B\) is the event where both A and B occur.
• Complement (\(A̅\)): \(A̅\) is the event where A does not occur.

For any two events A and B, the general formula for the probability of their union is \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). When events are mutually exclusive, \(P(A \cap B) = 0\), simplifying the formula to \(P(A \cup B) = P(A) + P(B)\).

The term \(A̅ \cap B\) represents the part of event B that does not overlap with event A. This is often written as \(B \setminus A\) (B minus A). For mutually exclusive events, there is no overlap (\(A \cap B = \emptyset\)), so the part of B that doesn't overlap with A is simply all of B (\(B \setminus A = B\)). This confirms why \(A̅ \cap B = B\) for mutually exclusive A and B.