P(At least two of A, B, C occures)
=$P(A\cap B\cap \bar{C})+P(A\cap\bar{B}\cap C)+P(\bar{A}\cap B\cap C)+P(A\cap B\cap C)$
Given that $P(A)=P(B)=P(C)$
$P(A\cap B\cap\bar{C})=P(A)P(B)P(\bar{C})$
$P\times P\times (1-P)$
=$P^{2}(1-P)$
$P(A\cap \bar{B}\cap C)=P(A)P(\bar{B})P(C)$
=$P(1-P)P$
=$P^{2}(1-P)$
$P(\bar{A}\cap B\cap C)=P(\bar{A})P(B)P(C)$
=$(1-P)P\times P$
$P(A\cap B\cap C)=P(A)P(B)P(C)$
=$P\times P\times P\times=P^{3}$
P(At least two of A, B, C occures)
=$P^{2}(1-P)+P^{2}(1-P)+P^{2}(1-P)+P^{3}$
=$3P^{2}-2P^{3}$
Hence given statement is True