permutations/combinatories and probabilityc

[shiven.bhandari]

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?

PS: I just made that problem up so source would be "me" :)

Cheers,

[ronaldramlan]

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,

[mschwrtz]

shiven.bhandari, your first method is the more elegant of the two.

Your first determine the answer as a product of probabilities, while your second counts outcomes, and plugs them into (favorable outcomes)/(total outcomes). In general, if you have can see a way to calculate your answer as a product of probabilities, you'll find less computation than if you have to do all that counting.

But ronaldramlan wass able to get through all that opportunity for compuational error. I'd like to clarify it for others who don't have quite his grasp of this area.

ronaldramlan's solution is to divide the favorable outcomes by the total outcomes. However, he uses this technique to determine the probability that neither chosen will play x. He then subtracts this probability from 1 to determine the value that at least one will

Favorable outcomes=the number of distinct pairs in a pool of 10 (often expressed as 10c2), or 10!/(8!2!).

Total outcomes = the number of distinct pairs in a pool of 15 (often expressed as 15c2), or 15!/(13!2!).

This yields 3/7, and 1 -(3/7)=4/7.

Finally, here's a third way: To determine the probability that the first person chosen will play x OR the second person chosen will play x, add the first probability (1/3) to the second probability (1/3), then subtract the probability that both people chosen will play x ((1/3)(2/7)). You need this last step because you will otherwise count some outcomes twice. If you have the MGMAT Strategy Guide 4, see the bottom of page 87 for more on this. (1/3) + (1/3) - (1/3)(2/7)= 4/7.