Question

An ordinary cube has 4 blank faces, one face marked 2 and another marked 3. Then the probability of obtaining a total of 9 in 5 throws is :

• A. 31/7776
• B. 5/2592
• C. 5/1944
• D. 5/162

Answer 1

Answer: (D)

$P(B)=\frac{4}{6}=\frac{2}{3};P($ face 2$)=\frac{1}{6}=P($ face 3$)$
$P($ total of 9 in five throws $)=P(33300$ or 22230$)$
$={\left(\frac{1}{6}\right)}^3\cdot {\left(\frac{2}{3}\right)}^2\cdot \frac{5!}{3!\cdot 2!}+{\left(\frac{1}{6}\right)}^3\cdot \frac{1}{6}\cdot \frac{2}{3}\cdot \frac{5!}{3!}=\frac{10}{216}\cdot \frac{4}{9}+\frac{1}{216}\cdot \frac{1}{9}\cdot 20$
$=\frac{1}{216.9}[60]=\frac{10}{9.36}=\frac{5}{9\cdot 18}=\frac{5}{162}$