Question

Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 4.

Answer 1

(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)

To Find: P(sum of faces neither divisible by 3 nor by 4) 

Sum = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} 

Sum neither divisible by 3 nor 4 = {2, 5, 7, 10, 11}

p =  \(\frac{number\,of\,favourable\,outcomes}{Total\,possible\,outcomes}\)

P(sum of faces neither divisible by 3 nor by 4) = \(\frac{5}{11}\)

Hence, probability is \(\frac{5}{11}\)