My maths Question is on Probability In an experiment a die is rolled twice, and the numbers shown are noted


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*what is the probability of getting the same number on the first and second   rolls?


my answer is $\frac1{36}$ but I just need confirmation on this cos im not that smart in probability 
 A: There are 6 possible outcomes for the first throw and 6 possible outcomes for the second throw, so you can model each event as a pair $(a, b) \in \{1, \dotsc, 6\}^2$.
Among the relevant events ("first number = second number") are the pairs $(2,2)$ and $(5,5)$. The pair $(6,5)$ is not.
What are all the relevant events? 
Your probability is the number of relevant events divided by the number of all possible events. (The result is different from the one you gave)
A: Just incase it's still not entirely clear, here, explicitly, is the set of every outcome you can get:
$$\{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), \dots \}$$
the questions essentially asks "How many times does $(x,x)$ occur in this set?". You responded with $\frac{1}{36}$, meaning that it only happens just once in this set of $36$ elements.
For the first roll you have a $\frac{1}{1}$ chance of getting just something. Call that something $x$. On your second roll, you have a $\frac{1}{6}$ chance of getting that same thing, $x$, again.
So that's why the answer is $$\dfrac{1}{1} \cdot \dfrac{1}{6} = \dfrac{1}{6}.$$
A: Your answer, $\frac{1}{36}$, is incorrect.
You can break this question down to: What is the probability to get the same number on the second role.
Also you can just write down every combination and count them for a correct answer.
For a good answer on probability questions, you also need to modell the experiment. 
