If \( P(A\cap B)=\displaystyle\frac{1}{2},\ P(A'\cap B')=\displaystyle\frac{1}{3},\ P(A)=p \) and \( P(B)=2p \), then value of \( p \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{7}{18} \)
Since \( A'\cap B'=(A\cup B)' \),
\( \displaystyle P(A\cup B)=1-\frac{1}{3} \)
\( \displaystyle =\frac{2}{3} \)
By the addition theorem, \( P(A\cup B)=P(A)+P(B)-P(A\cap B) \):
\( \displaystyle p+2p-\frac{1}{2}=\frac{2}{3} \)
\( \displaystyle 3p=\frac{7}{6} \)
\( \displaystyle p=\frac{7}{18} \)
Note: It seems that there is a mistake in the original NTA paper, which prints data that are not consistent: \( A\cap B \) is a part of A, so \( P(A) \) can never be smaller than \( P(A\cap B)=\frac{1}{2} \), yet the data give \( P(A)=p=\frac{7}{18} \), which is smaller than \( \frac{1}{2} \). The question can still be solved, because the addition theorem gives one definite value of p, and Option (4) is the only option equal to it.