The probability of getting exactly 1 six when 6 dice are rolled

$= (_6C_1)(1/6)^1(5/6)^5$

The probability of getting exactly *x* sixes when 6 dice are rolled

$= (_6C_x)(1/6)^x (5/6)^{6-x} ~ ~ ~ \text{where} ~ x = 0, \, 1, \, 2, \, 3, \, 4, \, 5, \, 6$

The probability of getting 1 or more sixes when 6 dice are rolled

$= 1 - (_6C_0)(1/6)^0 (5/6)^{6}$

The probability of getting exactly *x* sixes when *n* dice are rolled

$= (_nC_x)(1/6)^x (5/6)^{n - x} ~ ~ ~ \text{where} ~ x = 0, \, 1, \, 2, \, \ldots , \, n$

The probability of getting *n* or more sixes when 6*n* dice are rolled is

$\begin{align} \displaystyle

P & = \sum_{x = n}^{6n} ({_{6n}C_x})(1/6)^x (5/6)^{6n-x} \\

& = 1 - \sum_{x = 0}^{n - 1} ({_{6n}C_x})(1/6)^x (5/6)^{6n - x}

\end{align}$

For 6 dice, *n* = 1. The probability of getting 1 or more sixes is

$\begin{align} \displaystyle

P_a & = \sum_{x = 1}^{6} ({_6C_x})(1/6)^x (5/6)^{6-x} \\

& = 0.665

\end{align}$

For 12 dice, *n* = 2. The probability of getting 2 or more sixes is

$\begin{align} \displaystyle

P_b & = 1 - \sum_{x = 0}^{1} ({_{12}C_x})(1/6)^x (5/6)^{12 - x} \\

& = 0.619

\end{align}$

For 18 dice, *n* = 3. The probability of getting 3 or more sixes is

$\begin{align}

P_c & = 1 - \sum_{x = 0}^{2} ({_{18}C_x})(1/6)^x (5/6)^{18 - x} \\

& = 0.597

\end{align}$

*P*_{a} > *P*_{b} > *P*_{c}

The most likely to occur is (*a*) ← Answer: [ A ]

See also this problem at Wolfram post