Question 8 Exercise 6.5
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.
Question 8
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: $$=1-\dfrac{2}{9}=\dfrac{7}{9}$$
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