Question

If A and B are events such that \(P(A'UB')=\frac{1}{3}\) and \(P(AUB) = \frac{4}{9}\) then the value of \(P(A') + P(B')\) is:

• A. 1
• B. \(\frac{7}{9}\)
• C. \(\frac{8}{9}\)
• D. \(\frac{5}{9}\)

Answer 1

The Correct Option is C

We will use the complement rule and properties of probabilities to solve for \(P(A') + P(B')\).
Given: \(P(A' \cup B') = \frac{1}{3}\) and \(P(A \cup B) = \frac{4}{9}\).
The probability of the complement union can be expressed as:
\(P(A' \cup B') = 1 - P(A \cap B) = \frac{1}{3}\).
Using the formula for the union of any two events:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
From the information, we have:
\(P(A \cup B) = \frac{4}{9}\).
Substituting \(P(A' \cup B') = 1 - P(A \cap B) = \frac{1}{3}\) gives:
\(1 - P(A \cap B) = \frac{1}{3} \Rightarrow P(A \cap B) = \frac{2}{3}\).
Now using:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{4}{9}\),
\(P(A) + P(B) = \frac{4}{9} + \frac{2}{3} = \frac{4}{9} + \frac{6}{9} = \frac{10}{9}\).
Therefore, the sum of the probabilities of complements is:
\(P(A') + P(B') = 2 - (P(A) + P(B)) = 2 - \frac{10}{9} = \frac{18}{9} - \frac{10}{9} = \frac{8}{9}\).
Thus, the value of \(P(A') + P(B')\) is \(\frac{8}{9}\).