What is the probability of rolling an odd or non-prime number using a 12-sided die?

A) 1/2
B) 5/12
C) 1
D) 11/12

D) 11/12

Step-by-step explanation:

From 1 to 12, 6 numbers are odd (1, 3, 5, 7, 9, and 11).

7 numbers are non-prime (1, 4, 6, 8, 9, 10, and 12).

2 numbers are both odd and non-prime (1 and 9).

Therefore, the probability is:

(6 + 7 − 2) / 12 = 11/12

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option