Two dice are thrown simultaneously. Find the probability that the sum of numbers on both exceeds 8.

Two dice are thrown simultaneously. The probability that the sum of numbers on both exceeds 8 is 

(A) $\dfrac{4}{36}$ 

(B) $\dfrac{7}{36}$ 

(C) $\dfrac{9}{36}$ 

(D) $\dfrac{10}{36}$ 

(GATE 2005) 

Answer: (D) $\dfrac{10}{36}$ 

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, 

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

So, the number of event points, $n(E) = 10$. 

Therefore, the required probability is 

$= P(E) = \dfrac{n(E)}{n(S)} = \dfrac{10}{36}$.