Answer:
![P(A|B)=(2)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kz73xdwc1ztn0u4trlzfrzcemsw6agn4y8.png)
![P(A)*P(B)=(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vgrubbr452fxmb206ui3bcvjrzn4rwyeqd.png)
![P(A) =(2)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/odcfu5pb6ap4k3yx5xeb38o62783q0qwk1.png)
.
Explanation:
We use the Venn diagram to calculate the desired probabilities.
Note that there are 6 possible results in the sample space
S = {1, 2, 3, 4, 5, 6}
Then note that in the region representing the intercept of A and B there are two possible values.
So
![P (A\ and\ B) = (2)/(6) = (1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/w9s4f5rhonep6hb88tozjrdvxuwfxs6pw7.png)
In the region that represents event A there are 4 possible outcomes {4, 5, 1, 2}
So
![P(A) = (4)/(6) = (2)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/eqhtmc5q8xt6z4p8q0etnx0ykweqvl0sgl.png)
In the region that represents event B there are 3 possible outcomes {1, 2, 6}
So
.
Now
![P(A | B)=(P(A \ and\ B))/(P(B))\\\\P(A | B)=((1)/(3))/((1)/(2))\\\\P(A|B)=(2)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/991jzokbnl2vbm4zh0jy7siwszag8metsp.png)
![P(A)*P(B)=(2)/(3)*(1)/(2)=(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/fzgfjtbfi3529of6jo1drwsd426lpujnpm.png)