Let A B, and C be three events such that the probability that exactly one of A and B occurs is (1− k), the probability that exactly one of B and C occurs is (1 − 2k), the probability that exactly one of C and A occurs is (1− k) and the probability of all A B, and C occur simultaneously is k², where 0 < k < 1. Then the probability that at least one of A B, and C occur is

There are three events A, B, C.

Probability that exactly one of A and B occur is

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

Probability that exactly one of B and C occur is

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

Probability that exactly one of C and A occur is

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

Probability of all A, B, C occur simultaneously, is

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

From Eqs. (i), (ii) and (iii),

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

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

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

Adding Eqs. (v), (vi) and (vii),

$\begin{array}{l}2\left[\mathrm{P}\right(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A}\cap \mathrm{B})-\mathrm{P}(\mathrm{B}\cap \mathrm{C})-\mathrm{P}(\mathrm{C}\cap \mathrm{A}\left)\right]\\ =(1-\mathrm{k})+(1-2\mathrm{k})+(1-\mathrm{k})\\ =3-4\mathrm{k}\end{array}$

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

$\begin{array}{c}\Rightarrow \mathrm{P}(\mathrm{A}\cup \mathrm{B}\cup \mathrm{C})=\frac{3-4\mathrm{k}}{2}+\mathrm{P}(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C})\\ =\frac{3-4\mathrm{k}}{2}+{\mathrm{k}}^2=\frac{2{\mathrm{k}}^2-4\mathrm{k}+3}{2}\end{array}$

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

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