JEE Main 2021 — Probability Question with Solution

Given:

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

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

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

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

Now,

$P\left(A\right)+P\left(B\right)-2P(A\cap B)=1-k \dots \left(\mathrm{i}\right)$

$P\left(B\right)+P\left(C\right)-2P(B\cap C)=1-2k \dots \left(\mathrm{ii}\right)$

$P\left(C\right)+P\left(A\right)-2P(A\cap C)=1-k \dots \left(\mathrm{iii}\right)$

Adding $\left(i\right), \left(ii\right) \& \left(iii\right)$, we get

$P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(C\cap A\right)=\frac{-4k+3}{2}$

Now,

$P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(C\cap A\right)+P\left(A\cap B\cap C\right)$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{-4k+3}{2}+k^2$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{2k^2-4k+3}{2}$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{2(k-1{)}^2+1}{2}$

$\Rightarrow P\left(A\cup B\cup C\right)={\left(k-1\right)}^2+\frac{1}{2}$

$\Rightarrow P(A\cup B\cup C)>\frac{1}{2}$