Finding the Maximum Value of P(A) + P(B)

We are given information about two events, A and B. The problem provides constraints on the probability of their union (\(P(A \cup B)\)) and their intersection (\(P(A \cap B)\)). We need to find the maximum possible value for the sum of their individual probabilities, \(P(A) + P(B)\).

Understanding the Key Probability Formula

The relationship between the probabilities of the union and intersection of two events is fundamental in probability theory. The formula is:

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

We can rearrange this formula to express the sum \(P(A) + P(B)\):

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

Applying the Given Constraints

The problem states the following conditions:

• \(P(A \cup B) \ge 0.75\)
• \(0.125 \le P(A \cap B) \le 0.375\)

Our goal is to maximize the value of \(P(A) + P(B)\). Based on the rearranged formula, \(P(A) + P(B) = P(A \cup B) + P(A \cap B)\), we need to find the maximum possible values for both \(P(A \cup B)\) and \(P(A \cap B)\) that satisfy the given constraints.

Determining the Maximum Values

1. Maximum value for \(P(A \cup B)\): The probability of any event cannot exceed 1. Therefore, the maximum possible value for \(P(A \cup B)\) is 1. This satisfies the condition \(P(A \cup B) \ge 0.75\). 2. Maximum value for \(P(A \cap B)\): The problem explicitly gives the upper limit for the intersection probability: \(P(A \cap B) \le 0.375\). Thus, the maximum value for \(P(A \cap B)\) is \(0.375\).

Calculating the Maximum Sum

Now, we substitute these maximum values back into the rearranged formula:

\(\text{Maximum value of } (P(A) + P(B)) = (\text{Maximum value of } P(A \cup B)) + (\text{Maximum value of } P(A \cap B))\)

\(\text{Maximum value of } (P(A) + P(B)) = 1 + 0.375\)

\(\text{Maximum value of } (P(A) + P(B)) = 1.375\)

Verification

Let's check if this scenario is possible. If we set \(P(A \cup B) = 1\) and \(P(A \cap B) = 0.375\), then \(P(A) + P(B) = 1.375\). We also need to ensure that there exist probabilities \(P(A)\) and \(P(B)\) such that these conditions hold and \(P(A) \ge P(A \cap B)\) and \(P(B) \ge P(A \cap B)\). Consider \(P(A) = 1\) and \(P(B) = 0.375\). Then \(P(A) + P(B) = 1 + 0.375 = 1.375\). \(P(A \cap B) = P(A) + P(B) - P(A \cup B) = 1.375 - 1 = 0.375\). This configuration satisfies all the given conditions:

• \(P(A \cup B) = 1\), which is \(\ge 0.75\).
• \(P(A \cap B) = 0.375\), which is within the range \([0.125, 0.375]\).
• \(P(A)=1\) and \(P(B)=0.375\) are valid probabilities, and \(P(A) \ge P(A \cap B)\) and \(P(B) \ge P(A \cap B)\) are satisfied.

Therefore, the maximum value of \(P(A) + P(B)\) is \(1.375\).