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If \[P\left( {\overline A \cap B} \right) + P\left( {A \cap \overline B } \right) = 1 - k,\] \[P\left( {\overline A \cap C} \right) + P\left( {A \cap \overline C } \right) = 1 - 2k,\] \[P\left( {\overline B \cap C} \right) + P\left( {B \cap \overline C } \right) = 1 - k,\] $P\left( {A \cap B \cap C} \right) = {k^2},$ $k \in \left( {0,1} \right)$. Then, find the value of $P$ $($at least one of $A,B,C)$ is:
(A) $ > \;\dfrac{1}{2}$
(B) $\left[ {\dfrac{1}{8},\dfrac{1}{4}} \right]$
(C) \[ < \;\dfrac{1}{4}\]
(D) $\dfrac{1}{4}$

Answer
VerifiedVerified
162.3k+ views
Hint: In order to solve this question, we will first simplify all three given equations using the suitable identity. Next, we will add all the equations obtained by solving the given three equations. Further, we will substitute them in another identity and simplify it to get the desired correct answer.

Complete step by step Solution:
We are given that,
\[P\left( {\overline A \cap B} \right) + P\left( {A \cap \overline B } \right) = 1 - k\] ………………..equation $\left( 1 \right)$
\[P\left( {\overline A \cap C} \right) + P\left( {A \cap \overline C } \right) = 1 - 2k\] ………………..equation $(2)$
\[P\left( {\overline B \cap C} \right) + P\left( {B \cap \overline C } \right) = 1 - k\] ………………..equation $(3)$
We know that,
$P\left( {\overline X \cap Y} \right) = P\left( Y \right) - P\left( {X \cap Y} \right)$ ………………..equation $(4)$
Solving equations $\left( 1 \right),\left( 2 \right),\left( 3 \right)$ using equation $(4)$ we get,
$P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right)$ ………………..equation $\left( 5 \right)$
$P\left( A \right) + P\left( C \right) - 2P\left( {A \cap C} \right)$ ………………..equation $\left( 6 \right)$
$P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right)$ ………………..equation $\left( 7 \right)$
Adding equations $\left( 5 \right),\left( 6 \right),\left( 7 \right)$ we get,
$P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) = \dfrac{{\left( {3 - 4k} \right)}}{2}$ ………………..equation $\left( 8 \right)$
We know that,
\[P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)\]
Substituting equation $\left( 8 \right)$ in the above equation,
$P\left( {A \cup B \cup C} \right) = \dfrac{{\left( {3 - 4k} \right)}}{2} + {k^2}$
Solving it,
$P\left( {A \cup B \cup C} \right) = \dfrac{{\left( {3 - 4k + 2{k^2}} \right)}}{2}$
Since, the value of $2{k^2} - 4k + 3$ is greater than $1$ so,
$P\left( {A \cup B \cup C} \right) > \;\dfrac{1}{2}$

Hence, the correct option is A.

Note: Make sure you use a suitable identity. Also, after solving the given equations, in order to avoid getting a wrong answer, check wisely and attentively whether you need to add all the equations or some other simplification is required depending upon the need of the question.