Question:
If \( A \) and \( B \) are two events such that \( P(A\cup B)=\frac{5}{8} \) and \( P(A'\cup B')=\frac{3}{4} \), then \( P(A')+P(B')= \)
If \( A \) and \( B \) are two events such that \( P(A\cup B)=\frac{5}{8} \) and \( P(A'\cup B')=\frac{3}{4} \), then \( P(A')+P(B')= \)
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Use \( A'\cup B'=(A\cap B)' \) to get \( P(A\cap B) \), then the addition theorem to get \( P(A)+P(B) \).
Updated On: Nov 12, 2026
- \(\frac{1}{8}\)
- \(\frac{3}{8}\)
- \(\frac{9}{8}\)
- \(\frac{11}{8}\)
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The Correct Option is C
Solution and Explanation
By De Morgan's law, \( A'\cup B'=(A\cap B)' \), so \( P(A\cap B)=1-\frac{3}{4}=\frac{1}{4} \).
Addition theorem: \( P(A)+P(B)=P(A\cup B)+P(A\cap B)=\frac{5}{8}+\frac{1}{4}=\frac{7}{8} \).
Then \( P(A')+P(B')=2-[P(A)+P(B)]=2-\frac{7}{8}=\frac{9}{8} \).
Answer: (3) \( \frac{9}{8} \)
Addition theorem: \( P(A)+P(B)=P(A\cup B)+P(A\cap B)=\frac{5}{8}+\frac{1}{4}=\frac{7}{8} \).
Then \( P(A')+P(B')=2-[P(A)+P(B)]=2-\frac{7}{8}=\frac{9}{8} \).
Answer: (3) \( \frac{9}{8} \)
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