Solutions of Question 8 of Exercise 6.5 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
If pair ol dice is thrown, find the probability that the sum of digits is neither $7$ nor $11.$ Solution: The sample space rolling a pair of dice is \begin{align}s&=(i i, j): i, j-1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1.1) & (1.2) & (1.3) & (1.4) & (1.5) & (1.6) \\ (2.1) & (2.2) & (2.3) & (2.4) & (2.5) & (2.6) \\ (3.1) & (3.2) & (3.3) & (3.4) & (3.5) & (3.6) \\ (4.1) & (4.2) & (4.3) & (4.4) & (4.5) & (4.6) \\ (5.1) & (5.2) & (5.3) & (5.4) & (5.5) & (5.6) \\ (6.1) & (6.2) & (6.3) & (6.4) & (6.5) & (0.6) \end{array}\right]\end{align} The probability that the sum of number is $7$ is: $$=\dfrac{6}{36}$$ The probability that the sum of number is $11$ is: $$=\dfrac{2}{36}$$ The events that the sum is $7$ or $11$ are mutually exclusive,
therefore by addition law of probability we have
\begin{align}P(A \cup B)&=P(A)+P(B)\\ &=\dfrac{6}{36}+\dfrac{2}{36}\\ &=\dfrac{8}{36}\\ &=\dfrac{2}{9}\end{align} Thus by complementary events, the probability that sum of the digits is neither $7$ nor $11$ is: $$=1-\dfrac{2}{9}=\dfrac{7}{9}$$