axs wrote:
Thanks for a recapitulation of the combinations method for solving this problem.
Unfortunately, I still do not understand the critical step.
Yes, choosing any 2 from 6 (6C2) = 15.
But then you say "1C1 (= 1) is one way to pick 3 and 10". Huh?
It's not making sense to me.
If I say that I want to know the probability of choosing disc "3", I'm choosing 1 disk from among 10.
How is that a probability of 1?
Let me add something here:
The terms Combinations and Probability are related but different.
Probability (A) = No of suitable outcomes where A happens/Total no of outcomes
We use combinations to get "No of suitable outcomes where A happens" and "Total no of outcomes" and then calculate the required probability. So for the time being, just forget that you have to find some probability.
Focus on two things: "No of suitable outcomes where A happens" and "Total no of outcomes"
It is easy to find total number of outcomes. 10C4. e.g. (2, 3, 4, 6), (4, 1, 9, 5) etc etc etc
In how many combinations is range 7? (This is the "No of suitable outcomes where A happens")
(3, 10, a, b), (2, 9, c, d) and (1, 8, e, f)
a and b are numbers between 3 and 10 (so that range doesn't exceed 7)
c and d are numbers between 2 and 9
e and f are numbers between 1 and 8
In how many ways can you choose a and b? 6C2 (Any 2 out of 6 numbers between 3 and 10). So how many 4 number combinations are there which look like this: (3, 10, 4, 6), (3, 10, 7, 5) etc?
I hope you agree it is 6C2.
Similarly, you get 6C2 combinations of the type (2, 9, c, d) and 6C2 of the type (1, 8, e, f).
No of suitable combinations = 6C2 + 6C2 + 6C2
Now, the required probability = No of suitable combinations where A happens/Total no of outcomes
= (3*6C2)/10C4