A single die is thrown two times. What is the probability that the sum is neither 8 nor 9?

A single die is thrown two times. What is the probability that the sum is neither 8 nor 9? 

(A) $\dfrac{1}{9}$ 

(B) $\dfrac{5}{36}$ 

(C) $\dfrac{1}{4}$ 

(D) $\dfrac{3}{4}$ 

(GATE 2005) 

Answer:  (D) $\dfrac{3}{4}$ 

Explanation: 

Here, the sample space, 

$S = \{(1, 1), \cdots, (1, 6), (2, 1), \cdots, (2, 6), (3, 1), \cdots, \\(3, 6), (4, 1), \cdots, (4, 6), (5, 1), \cdots, (5, 6), (6, 1), \cdots, (6, 6) \}.$ 

So, the number of sample points, $n(S) = 6 \times 6 = 36$. 

Again, favorable event space, 

$E = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), \\~~~~~~~~~~(3, 6), (4, 5), (5, 4), (6, 3)\}.$  

So, the number of event points, $n(E) = 9$. 

Therefore, the probability that the sum is either 8 or 9 is 

$= P(E) = \dfrac{n(E)}{n(S)} = \dfrac{9}{36} = \dfrac{1}{4}$. 

Therefore, the required probability is 

$= 1 - P(E) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.