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.
Solution:
Given: Two dice are thrown together.
Sample Space:
$(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)$
$(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)$
$(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)$
$(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)$
$(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)$
$(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$
To Find: $\mathrm{P}$ (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 $4=\{2,5,7,10,11\}$
$P=\frac{\text { number of favourable outcomes }}{\text { total possible outcomes }}$
$P($ sum of faces neither divisible by 3 nor by 4$)=\frac{5}{11}$
Hence, probability is $\frac{5}{11}$.