Answer to Question #218797 in Statistics and Probability for King

Question

(a) Gideon rolls two (2) six-sided dice once. What is the probability that the sum of the  outcomes of both dice are odd or divisible by 5? (b) Two (2) dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on two different numbers?  

Answer 1

(a)

The sum of the outcomes of both dice

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hdashline 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hdashline 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hdashline 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hdashline 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hdashline 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hdashline \end{array}$$

Let $O$ represents odd and $D$ represents divisible by 5.

$$N(O\cup D )=N(O)+N(D)-N(O\cap D)$$

$$=18+7-4=21$$

$$P(O\cup D)=\dfrac{21}{36}=\dfrac{7}{12}$$

(b) Two different numbers

$$(1, 2), (1,3), (1, 4), (1, 5), (1,6),$$

$$(2, 1), (2,3), (2, 4), (2, 5), (2,6),$$

$$(3, 1), (3,2), (3, 4), (3, 5), (3,6),$$

$$(4, 1), (4,2), (4, 3), (4, 5), (4,6),$$

$$(5,1), (5,2), (5,3), (5,4), (5,6),$$

$$(6, 1),(6, 2), (6,3), (6, 4), (6, 5).$$

$$P(6|Different)=\dfrac{10}{30}=\dfrac{1}{3}$$