Question

A pair of dice is thrown. Find the probability of getting a sum of numbers which is perfect cube and divisible by $3$.

Answer 1

When a pair of dice are thrown then total no. of outcome$=36$

Outcomes:

[(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), (6,6)]

Now the possible sum of numbers which is perfect cube and divisible by $3$ is $=0$

So, net probability $=\frac{0}{16}=0$