I have a maths exam tomorrow and this these questions are bugging me.
Question 1: Two numbers are randomly selected from set of first 6 natural numbers, find probability that selected numbers are co-prime.
Answer 1: Favorable cases are 11, but total cases are only 6C2, which is fine because if we pick 1,6 and 6,1, they are the same thing.
Question 2: Two dices are thrown simultaneously, what is the probability that the the sum of number is 6
Answer 2: Now, in this question why do we treat (2,4), and (4,2) as 2 different cases when in the above question we don’t?
I know this kind of sounds obvious but I a really confused about this, and would really appreciate some help
It depends on what your basic "event" is. In your case,
Note that:
For answer 1, you could also take $(1,6)$ and $(6,1)$ as two different events. However, in that case, there are more than $11$ favorable cases. In fact, there are now $22$ favorable cases, and the probability you end up is the same, except you had to double count some things, overall complicating the solution. Taking unordered sets was the least complicated method.
For answer 2, you could also take unordered pairs as your base events. However, in that case, the events $\{1,1\}$ and $\{1,2\}$ would no longer have equal probability, so that would just complicate solving the final problem! Taking ordered pairs was the least complicated method.