Question

If three six faced die each marked with numbers $1$ to $6$ on six faces, are thrown find the total number of possible outcomes that atleast one dice shows the number 2?

Answer 1

Number of required ways $=$ Total number of Outcomes - Number of outcomes without 2's

$=6\ast 6\ast 6-5\ast 5\ast 5=6^3-5^3=91$