Let A, B and C be three events such that the probability that...

Given:

$P ( \overset{¯}{A} \cap B ) + P ( A \cap \overset{¯}{B} ) = 1 - k$

$P ( \overset{¯}{A} \cap C ) + P ( A \cap \overset{¯}{C} ) = 1 - k$

$P ( \overset{¯}{B} \cap C ) + P ( B \cap \overset{¯}{C} ) = 1 - 2k$

$P ( A \cap B \cap C ) = {k}^{2}$

Now,

$P ( A ) + P ( B ) - 2P ( A \cap B ) = 1 - k \ldots \left(i\right)$

$P ( B ) + P ( C ) - 2P ( B \cap C ) = 1 - 2k \ldots \left(ii\right)$

$P ( C ) + P ( A ) - 2P ( A \cap C ) = 1 - k \ldots \left(iii\right)$

Adding \left(i\right), \left(ii\right) \text{ and } \left(iii\right)$, we get</p><p>$P\left(A\right) + P\left(B\right) + P\left(C\right) - P\left(A \cap B\right) - P\left(B \cap C\right) - P\left(C \cap A\right) = \dfrac{- 4k + 3}{2}$</p><p>Now,</p><p>$P\left(A \cup B \cup C\right) = P\left(A\right) + P\left(B\right) + P\left(C\right) - P\left(A \cap B\right) - P\left(B \cap C\right) - P\left(C \cap A\right) + P\left(A \cap B \cap C\right)$</p><p>$⇒ P\left(A \cup B \cup C\right) = \dfrac{- 4k + 3}{2} + {k}^{2}$</p><p>$⇒ P\left(A \cup B \cup C\right) = \dfrac{2{k}^{2} - 4k + 3}{2}$</p><p>$⇒ P\left(A \cup B \cup C\right) = \dfrac{2 ( k - 1{)}^{2} + 1}{2}$</p><p>$⇒ P\left(A \cup B \cup C\right) = {\left(k - 1\right)}^{2} + \dfrac{1}{2}$</p><p>$⇒ P ( A \cup B \cup C ) > \dfrac{1}{2}$