In a stack of six cards, each card is labeled with a different integer 0 through 5. If **two **cards are selected at random without replacement, what is the probability that their sum will be 3?
How do I determine if order matters or doesn’t? Also, if order matters, would the probability be bigger or smaller than if the order doesn’t matter?
As YOU think about the problem, you get to choose whether you want order to matter or not. And as long as you are consistent in your choice, you will get the same answer either way.
Either way, you have to ask “how many possible outcomes are there?” and “Of those, how many have a sum of 3?”
It’s your choice whether you want to consider ordered pairs or just 2-number combinations, not counting order. There are more ordered pairs than there are 2-number combinations (in fact twice as many). But then you have to count 1,2 and 2,1 as distinct possibilities…Meanwhile, someone else who just chose to count un-ordered combinations will have fewer of them to deal with, but will count the 1,2 combination just once, not distinct from 2,1.
Hope that helps. I’ll add one more hint: like most SAT counting problems, this one is vulnerable to brute force. Just list out all of the possible combinations and then count how many have a sum of 3. Not elegant, but it takes like 30 seconds and it pretty safe.
So is the answer 2/25?
It’s 2/15.
So pckeller, you are thinking it like 4/30 and 2/15, right (a proportion)?
2/15.
Think of it more like, if the order is relevant, then there are twice as many possible outcomes and twice as many “good” outcomes, but the ratio is still the same.
Any of 6 cards can be chosen first. After that, there are five possibilities for a second card. So there are 5 possible outcomes in step 2 (choosing 2nd card) for each of the 6 possible outcomes in step 1 (choosing first card) or 30 possible outcomes. If any of 0-3 are chosen first there will be one possibility ( a 1/5 chance) of the 2 cards having a sum of 3. So there are a total of 4 possibilities for a sum of 3 out of a total of 30 possibilities, so 4/30 or 2/15.
This must be an old problem – I just remembered an alternative solution that I remember liking when I first saw it a good while back. Actually, @CHD2013 's last post almost does it this way:
Say you are drawing the cards, hoping to come up with a winning sum. On your first pick, you are ok as long as from the 6 cards you start with, you pick the 0, 1, 2, or 3 – that’s a 4/6 probability that you are on your way to a win. Then, on the second pick, if you are still in the running, of the 5 remaining cards there is only one that will complete your sum to the desired value. So that’s a 1/5 probability.
The overall probability of success is then: (4/6) x (1/5) = 4/30 = 2/15.
I think it’s a slick answer and it does not use any counting methods.