Answer to Question #111194 in Statistics and Probability for Nimra

Question

Two fair dice are rolled. Find the probability N ?

The sum of 7.

Sum more then 8.

Sum of at least 8.

Sum of almost 4.

One phase show no 5 and other phase show less than 5 .

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c} (1,1) & (1,2) & (1,3) & (1,4) & (1, 5) & (1,6) \\ \hline (2,1) & (2,2) & (2,3) & (2,4) & (2, 5) & (2,6)\\ \hline (3,1) & (3,2) & (3,3) & (3,4) & (3, 5) & (3,6) \\ \hline (4,1) & (4,2) & (4,3) & (4,4) & (4, 5) & (4,6) \\ \hline (5,1) & (5,2) & (5,3) & (5,4) & (5, 5) & (5,6) \\ \hline (6,1) & (6,2) & (6,3) & (6,4) & (6, 5) & (6,6) \end{array}$$

The sum of 7.

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

$$P(sum \ of\ 7)={6 \over 36}={1 \over 6}$$

Sum more than 8.

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

$$P(sum \ more\ than \ 8)={10 \over 36}={5 \over 18}$$

Sum at least 8. 

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

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

$$P(sum \ at\ least \ 8)={15 \over 36}={5 \over 12}$$

Sum at most 4. 

$(1, 1), (1, 2), (1,3), (2, 1), (2, 2), (3, 1)$

$$P(sum \ at\ most\ 4)={6 \over 36}={1 \over 6}$$

One phase show no 5 and other phase show less than 5 .

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

$$P(one \ =5,\ other\ <5)={8 \over 36}={2 \over 9}$$