[tex]P(A)=\dfrac{|A|}{|\Omega|} [/tex]
[tex]\Omega=\{1;\ 2;\ 3;\ 4;\ 5;\ 6\}\to|\Omega|=6[/tex]
The first roll:
[tex]A_1=\{1;\ 3;\ 5\}\to|A_1|=3\\\\P(A_1)=\dfrac{3}{6}=\dfrac{1}{2}[/tex]
The second roll:
[tex]A_2=\{1;\ 2\}\to|A_2|=2\\\\P(A_2)=\dfrac{2}{6}=\dfrac{1}{3}[/tex]
Probability of the situation:
[tex]P(A)=P(A_1)\cdot P(A_2)\\\\P(A)=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}[/tex]
Answer: [tex]\boxed{\dfrac{1}{6}}[/tex]
