Question

Please help please help please

Answer 1

`P(A)=(|A|)/(|\Omega|)`


`\Omega=\{1;\ 2;\ 3;\ 4;\ 5;\ 6\}\to|\Omega|=6`

The first roll:


`A_1=\{1;\ 3;\ 5\}\to|A_1|=3\\\\P(A_1)=(3)/(6)=(1)/(2)`

The second roll:


`A_2=\{1;\ 2\}\to|A_2|=2\\\\P(A_2)=(2)/(6)=(1)/(3)`

Probability of the situation:


`P(A)=P(A_1)\cdot P(A_2)\\\\P(A)=(1)/(2)\cdot(1)/(3)=(1)/(6)`

Answer:
`\boxed{(1)/(6)}`