Question

Three red and three blue flags are arranged randomly along

Answer 1

$|\Omega|=\dfrac{6!}{3!3!}=\dfrac{4\cdot5\cdot6}{2\cdot3}=20\\A=\{RBRBRB,BRBRBR\}\\|A|=2\\\\P(A)=\dfrac{2}{20}=\dfrac{1}{10}$

Answer 2

The answer would be B