Question

A dice is thrown twice. Find the probability of getting a bigger value on the first throw.

Answer 1

There are 36 outcomes for rolling a die twice.
When two dices are thrown, the list of all possible outcomes are:
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Let us first consider the event “Getting the same value” for rolling a die twice. There are 6 favourable outcomes, so the probability is $\frac{6}{36}$.
So, the probability of getting different values on rolling a die twice $=$ 1 - $\frac{6}{36}$ = $\frac{30}{36}$
Of these, getting a bigger value on the first and getting a bigger value on the second are equally likely.
Hence, both these probabilities = $\frac{1}{2}\times \frac{30}{36}=\frac{15}{36}$