Question

An ordinary cube has four blank faces, one face marked $2$ another marked $3$. Then the probability of obtaining a total of exactly $12$ in $5$ throws is

• A. $\frac{5}{1296}$
• B. $\frac{5}{1944}$
• C. $\frac{5}{2592}$
• D. None of these

Answer 1

The correct option is C

$\frac{5}{2592}$

Explanation for the correct option:

Step 1. Find the required probability:

Given, a cube has 4 blank faces, one face marked as $2$ and the other face marked as $3$.

A cube is thrown five times. So, the total number of cases $n$$=6^5$

Step 2. A total of $12$ in$5$ throws can be obtained in the following two ways only:

(i) One blank and four $3\text{'}s$

(ii) Three $2\text{'}s$ and two $3\text{'}s$

The number of ways in case (i) is $={}^{5}C_1$

$\Rightarrow$ ${}^{5}C_1=\frac{5!}{1!(4!)}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$\Rightarrow$ ${}^{5}C_1=\frac{5\times 4!}{4!}$

$\Rightarrow$ ${}^{5}C_1=5$

and the number of ways in case (ii) is ${=}^5{C}_2$

$\Rightarrow$ ${}^{5}C_2=\frac{5\times 4\times 3!}{2!\times 3!}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$\Rightarrow$ ${}^{5}C_2=\frac{5\times 4}{2\times 1}$

$\Rightarrow$ ${}^{5}C_2=\frac{20}{2}$

$\Rightarrow$ ${}^{5}C_2=10$

Step 3. Let $m=$the favorable number of ways.

$$\begin{gathered} =5+10 \\ =15 \end{gathered}$$

$\therefore$the required probability$=\frac{15}{6^5}$ $\left[\because Probability=\frac{Favorablecases}{Totalnumberofcases}\right]$

$=\frac{\mathbf{5}}{\mathbf{2592}}$

Hence option(C) is correct.