Question

Please help with both Step by Steps!!!!

Answer 1

Rolling two dice can output any number between 2 and 12. Here are the probabilities (given by how many combinations would yield that sum):

You get 2 if you roll (1, 1) $\implies P=\dfrac{1}{36}$

You get 3 if you roll (1, 2), (2, 1) $\implies P=\dfrac{2}{36}$

You get 4 if you roll (1, 3), (2, 2), (3, 1) $\implies P=\dfrac{3}{36}$

You get 5 if you roll (1, 4), (2, 3), (3, 2), (4, 1) $\implies P=\dfrac{4}{36}$

You get 6 if you roll (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) $\implies P=\dfrac{5}{36}$

You get 7 if you roll (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) $\implies P=\dfrac{6}{36}$

You get 8 if you roll (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) $\implies P=\dfrac{5}{36}$

You get 9 if you roll (3, 6), (4, 5), (5, 4), (6, 3) $\implies P=\dfrac{4}{36}$

You get 10 if you roll (4, 6), (5, 5), (6, 4) $\implies P=\dfrac{3}{36}$

You get 11 if you roll (5, 6), (6, 5) $\implies P=\dfrac{2}{36}$

You get 12 if you roll (6, 6) $\implies P=\dfrac{1}{36}$

So, you get an even sum OR a 3 if you roll a total of 2, 3, 4, 6, 8, 10 or 12. The sum of their probabilities is

$\dfrac{1+2+3+5+5+3+1}{36}=\dfrac{20}{36}=\dfrac{5}{9}$

You get doubles and at least 8 if you get (4,4), (5,5) or (6,6). Each of these pairs happen with probability 1/36, for a total of

$\dfrac{3}{36}=\dfrac{1}{12}$