Question

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

• A. $\frac{4}{36}$
• B. $\frac{7}{36}$
• C. $\frac{9}{36}$
• D. $\frac{10}{36}$

Answer 1

The correct option is D $\frac{10}{36}$

Sample space S= {(1, 1), (1, 2), (1, 3), .........., (6, 6)} = 36 outcomes

Let E = {sum of Numbers > 8}

Then E = {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)} = 10 possibilities

$HenceP\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{36}$