If \(P(A)=\displaystyle\frac{1}{2}\), \(P(B)=\displaystyle\frac{1}{3}\), and \(P(A\cap B)=\displaystyle\frac{1}{15}\), then value of \(P(A'\cap B')\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{7}{30}\)
By De Morgan’s law, \(A'\cap B' = (A\cup B)'\), so
\(P(A'\cap B') = 1-P(A\cup B)\)
By the addition rule:
\( \displaystyle P(A\cup B) = \frac{1}{2}+\frac{1}{3}-\frac{1}{15} \)
\( \displaystyle = \frac{15+10-2}{30} = \frac{23}{30} \)
Hence
\( \displaystyle P(A'\cap B') = 1-\frac{23}{30} = \frac{7}{30} \)