Question

1. A pair of fair dice is tossed. Find the probability of getting
a. A total of 6.
b. At most a total of 6.

Answer 1

$$\begin{matrix} & 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 1+1 & 1+2 & 1+3 & 1+4 & 1+5 & 1+6 \\ 2 & 2+1 & 2+2 & 2+3 & 2+4 & 2+5 & 2+6\\ 3 & 3+1 & 3+2 & 3+3 & 3+4 & 3+5 & 3+6 \\ 4 & 4+1 & 4+2 & 4+3 & 4+4 & 4+5 & 4 +6 \\ 5 & 5+1 & 5+2 & 5+3 & 5+4 & 5+5 & 5+6 \\ 6 & 6+1 & 6+2 & 6+3 & 6+4 & 6+5 & 6+6 \end{matrix}$$

Total number of cases when two dice roles =6×6=36

Let us define the event A as:

a. A: On throw of the pair of fair dice, getting a total of 6

$$A=\{(1, 5), (2,4), (3,3),(4,2), (5,1)\}$$

$$P(a\ total\ of \ 6)=P(A)=\dfrac{5}{36}$$

b. Let us define the event B as:

B: On throw of the pair of fair dice, getting at most a total of 6

$$B=\{(1,1), (1,2),(1,3), (1,4),(1, 5),$$ $$(2,1), (2,2), (2,3), (2,4),$$ $$(3,1),(3,2), (3,3),$$ $$(4,1),(4,2),$$ $$(5,1)\}$$ $$P(at\ most\ a\ total\ of \ 6)=P(B)=\dfrac{15}{36}=\dfrac{5}{12}$$