Question

An ordinary cubical die has four blank faces, one face marked 2 and another marked 3. Then the probability of obtaining 12 in 5 throws is

• A. $\frac{5}{1296}$
• B. $\frac{5}{1944}$
• C. $\frac{5}{2529}$
• D. $\frac{5}{1264}$

Answer 1

The correct option is A $\frac{5}{1296}$

$\begin{array}{cccc}cases:& 2& 3& blanks\\ Case1:& -& 4& 1\\ Cases2:& 3& 2& 0\end{array}$
$P\left(case1\right)=5C_4{\left(\frac{1}{6}\right)}^4\frac{4}{6}=\frac{20}{6^5}andP\left(case2\right)=5C_3{\left(\frac{1}{6}\right)}^2{\left(\frac{1}{6}\right)}^3=\frac{10}{6^5}$
$\therefore P\left(sum12\right)=P\left(case1\right)+P\left(case2\right)=\frac{30}{6^5}=\frac{5}{1296}$