shiven.bhandari wrote:
Hello,
I am a new kid on the block - my apologies if the category of this question does not belong to this thread.
I am struggling to understand in what cases do we need the knowledge of combinations/permutations in order to calculate Probabilty. For example, consider this problem
Total players = 15; 5 play X, 3 play Y and rest Z. Two players are chosen at random. Probability that neither play X.
One way to solve is
P[no X] = 10/15 * 9/14
Another way is
5C2/15C2.
Which one is right and why? Basically, whichever way you choose should give you the same result. However, (10/15 * 9/14) and (5C2/15C2) won't give you the same result.
First, we know that 5 people play X, 3 people play Y and 7 people play Z. Then, we also know that P(choosing 2 people and neither plays X) = n(choosing 2 people out of those who play Y and Z) / n(all possible outcome of choosing 2 people)
You will get the result of "n(choosing 2 people out of those who play Y and Z)" from (3+7)C2 = 10C2, and I can see that you already know that the result of "n(all possible outcome of choosing 2 people)" is 15C2. Therefore, the probability of choosing 2 players and that none of them plays X is 10C2/15C2.
PS: I just made that problem up so source would be "me" :)
Cheers,