Thank you for reporting, we will resolve it shortly
Q.
A die is thrown three times, the probability of getting a larger number than the previous number each time, is
Probability
Solution:
Total number of exhaustive events $=6^{3}=216$.
Now in the case of die $2>1$ (i.e. second number is larger than first number). Let second number is $i(i>1)$ then the first number can be selected by / - 1 ways and third number by $6-i$ ways so the number of ways in which three numbers can be choosen are $(i-1) \times 1 \times(6-i)$ So, the total number of favourable cases are
$$
=\sum_{i=2}^{5}(i-1)(6-i)=20
$$
Hence, required probability is $=\frac{20}{216}=\frac{5}{54}$