If \( P(A \cap B) + P(B \mid A \cap B) = \), then:
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When dealing with conditional probabilities, remember that \( P(B \mid A \cap B) \) simplifies to 1 if the event \( A \cap B \) is a subset of \( B \).
We are given the expression: \[ P(A \cap B) + P(B \mid A \cap B), \] and we are asked to find its value. ### Step 1: Break down the components. We know that \( P(B \mid A \cap B) \) is the conditional probability of event \( B \) occurring given that \( A \cap B \) has occurred. By the definition of conditional probability, we have: \[ P(B \mid A \cap B) = \frac{P(B \cap A \cap B)}{P(A \cap B)}. \] Since \( A \cap B \) is the intersection of \( A \) and \( B \), we can simplify the expression: \[ P(B \mid A \cap B) = 1, \] because \( B \cap A \cap B = A \cap B \) by the properties of intersections. ### Step 2: Simplify the given expression. Substitute \( P(B \mid A \cap B) = 1 \) into the original expression: \[ P(A \cap B) + P(B \mid A \cap B) = P(A \cap B) + 1. \] Thus, the expression simplifies to: \[ P(A \cap B) + 1 = 2. \] ### Step 3: Conclusion. The final value of the given expression is \( 2 \). Thus, the correct answer is: \[ \boxed{2}. \]
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Approach Solution -2
To solve the question, let's start by analyzing the given expression \( P(A \cap B) + P(B \mid A \cap B) \). We need to evaluate and simplify this expression to determine its value. First, recall the definition of conditional probability:
\[ P(B \mid A \cap B) = \frac{P(A \cap B \cap B)}{P(A \cap B)} \]
Since \( A \cap B \cap B = A \cap B \), we have:
\[ P(B \mid A \cap B) = \frac{P(A \cap B)}{P(A \cap B)} = 1 \]
Thus, the expression \( P(A \cap B) + P(B \mid A \cap B) \) becomes:
\[ P(A \cap B) + 1 \]
Which simplifies to:
\[ P(A \cap B) + 1 = 2 \]
Therefore, the value of the given expression is \( 2 \), which corresponds to the correct answer:
