NIMCET 2017 Question Paper with solutions

We have,
P $\stackrel{\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}}{\left(\mathit{A}\cup \mathit{B}\right)}$ = $\frac{1}{6}$
P (A ∩ B) = $\frac{1}{4}$
P ($\stackrel{-}{\mathit{A}}$) = $\frac{1}{4}$
Now , P($\stackrel{-}{\mathit{A}}$) = $\frac{1}{4}$
⇒ 1 - P(A) = $\frac{1}{4}$
⇒ P (A) = $\frac{3}{4}$
Again P $\stackrel{\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}}{\left(\mathit{A}\cup \mathit{B}\right)}$ = $\frac{1}{6}$
⇒ 1 - P (A ∪ B) = $\frac{1}{6}$
⇒ P (A ∪ B) = 1 - $\frac{1}{6}$ = $\frac{5}{6}$
Now, we know that
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
⇒ $\frac{5}{6}$ = $\frac{3}{4}$ + P (B) - $\frac{1}{4}$
⇒ P (B) = $\frac{5}{6}-\frac{1}{2}$ = $\frac{1}{3}$
Again , P (A) P (B) = $\frac{3}{4}\times \frac{1}{3}$ = $\frac{1}{4}$ = P (A ∩ B)
and P (A) ≠ P (B)
∴ A and B are independent but not equally likely.