Question

A die is rolled three times, find the probability of getting every time a number larger than the previous number.

• A. $\frac{5}{54}$
• B. $\frac{7}{54}$
• C. $\frac{5}{34}$
• D. $\frac{7}{34}$

Answer 1

The correct option is A $\frac{5}{54}$

Let $S$ be the sample space and $E$ be the event space of getting a larger number larger than the previous number.

$\therefore n\left(S\right)=6\times 6\times 6=216$

Now we count the number of favourable ways.

Obviously, the second number has to be greater than $1^{st}.$

If the second number is $i(i>1),$ then the number of favourable ways $=(i=1)\times (6-i)$

$\therefore n\left(E\right)=$ Total number of favourable ways

$=\sum _{i=1}^6(i-1)\times (6-i)$

$=0+1\times 4+2\times 3+3\times 2+4\times 1+5\times 0$

$=4+6+6+4=20$

Therefore, the required probability is,

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{20}{216}=\frac{5}{54}$