Respuesta :
[tex]|\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}[/tex]
[tex]|\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}[/tex]