Question

A die is rolled three times. The probability of getting a larger number than the previous number each time is

• A. $\frac{5}{72}$
• B. $\frac{5}{54}$
• C. $\frac{13}{16}$
• D. $\frac{1}{18}$

Answer 1

The correct option is B

$\frac{5}{54}$

Find the probability of getting a larger number than the previous number each time.

According to the question, we have

$n=6,r=3$

Total number of sample space when die is rolled 3 times $=6^3$ [$\because$Total number of sample space when die is rolled $r$ times $={\mathbf{n}}^{\mathbf{r}}$]

Number of favourable outcomes $={}_{}{}^{6}C_3$

Therefore, the required probability $=\frac{{}_{}{}^{6}C_3}{6^3}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{P}\mathbf{r}\mathbf{o}\mathbf{b}\mathbf{a}\mathbf{b}\mathbf{i}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{N}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{f}\mathbf{a}\mathbf{v}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{a}\mathbf{b}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{s}\mathbf{}}{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{s}}\mathbf{]}$

$$\begin{gathered} =\frac{6!}{{\displaystyle \frac{\left(6-3\right)!3!}{216}}} \\ =\frac{5}{54} \end{gathered}$$

Hence, Option B is the correct answer.