Let A and B be two events such that P(B ∣ A) = (2)/(5) , P(A ∣ B) =...

Given,

$A$ and $B$ be two events such that $P\left(B ∣ A\right) = \dfrac{2}{5}$, $P\left(A ∣ B\right) = \dfrac{1}{7}$ and $P\left(A \cap B\right) = \dfrac{1}{9}$,

Now using the formula of conditional probability we get,

$P\left(A ∣ B\right) = \dfrac{1}{7} ⇒ \dfrac{P\left(A \cap B\right)}{P\left(B\right)} = \dfrac{1}{7}$

$⇒ \dfrac{\dfrac{1}{9}}{P\left(B\right)} = \dfrac{1}{7}$

$⇒ P\left(B\right) = \dfrac{7}{9}$

Now again using formula we get,

$P\left(B ∣ A\right) = \dfrac{2}{5} ⇒ \dfrac{P\left(A \cap B\right)}{P\left(A\right)} = \dfrac{2}{5}$

$⇒ \dfrac{\dfrac{1}{9}}{P\left(A\right)} = \dfrac{2}{5}$

$⇒ P\left(A\right) = \dfrac{5}{18}$

Now we know that,

$P\left({A}^{'} \cup B\right) = 1 - P\left(A \cup B\right) + P\left(B\right)$

$⇒ P\left({A}^{'} \cup B\right) = 1 - P\left(A\right) + P\left(A \cap B\right) = \dfrac{5}{6}$

And $P\left({A}^{'} \cap {B}^{'}\right) = 1 - P\left(A \cup B\right)$

{also $P\left(A \cup B\right) = P$\left(A\right) + P\left(B\right) - P\left(A \cap B\right)}</p><p>⇒ P\left({A}^{'} \cap {B}^{'}\right) = 1 - P\left(A\right) - P\left(B\right) + P\left(A \cap B\right) = \dfrac{1}{18}$</p><p>$⇒$Both \left(S1\right)$ and \left(S2\right) are true.