What Is the Probability of Selecting Two Cards That Sum to 3 from a Stack of 6?

[UrbanXrisis]

In a stack of 6 cards, each card is labled 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?

here's what I did... 0 1 2 3 4 5

to get either a 0,1,2, or 3 there is a 4/6 chance
after, there is a 1/5 chance to the the card that will add up to 3.

so...is the answer (4/6+1/5)=13/15?

I know that it's wrong but I'm not sure how to solve it

thanks,
Urban

 

[AKG]

There are 4 ways to choose two cards that sum to 3 ({1,2}, {2,1}, {0,3}, and {3,0}) and 6 x 5 = 30 ways to choose two cards. So, the probability is:

4/30 = 2/15

Your approach was also correct, but you should have multiplied the probabilities, not added them. It should be clear that adding doesn't make sense. Let's say you had 2 coins and are asked the probability that they're both heads. Adding gives you 1/2 + 1/2 = 1. Obviously that's wrong. Try it with 3 coins, you get a probability of 3/2, now that's obviously wrong. Now why should you multiply? Well, basically, you start off with some event, and you know the first part of the event only happens 4 out of 6 times. Now, even in the case where the first part goes right, the second part only goes right 1 out of 5 times. So, starting with 4 out of 6, only one fifth of those 4 will actually be any good. Hopefully you can see why you have to multiply (it seems kind of tough to put into words).

 

[loseyourname]

In a stack of 6 cards, each card is labled 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?

Let n = 6 = the number of cards in the stack.

Let r = 2 = the number of cards pulled.

Let N = the number of ways in which to pull two cards.

Let N(A) = the number of ways in which to pull two cards whose sum is 3.

$$N = C^{6}_{2} = \frac{6!}{2! 4!} = 15$$

We've already established that N(A) = 4 by listing out the ways.

So $$P(A) = \frac{N(A)}{N} = \frac{4}{15}$$

 

[Doc Al]

$$N = C^{6}_{2} = \frac{6!}{2! 4!} = 15$$

This assumes order doesn't matter.

We've already established that N(A) = 4 by listing out the ways.

If order didn't matter, the list would have 2 elements, not four.

AKG's analysis is correct.

 

[Didd]

We have six card,
1-card 2-card 3-card 4-card 5-card 6-card
0 1 2 3 4 5

+ 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 6
2 2 3 4 5 6 7
3 3 4 5 6 7 8 The experment have 25 out comes. But the event that
4 4 5 6 7 8 9 sum of two selected card is 3 will be 4 .
5 5 6 7 8 9 10 so probability=event/outcome = 4/25

 

[Gokul43201]

We have six card,
1-card 2-card 3-card 4-card 5-card 6-card
0 1 2 3 4 5

+ 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 6
2 2 3 4 5 6 7
3 3 4 5 6 7 8 The experment have 25 out comes. But the event that
4 4 5 6 7 8 9 sum of two selected card is 3 will be 4 .
5 5 6 7 8 9 10 so probability=event/outcome = 4/25

Didd, I think you mean to say 6*6 = 36 outcomes (not 25).

Anyway, this is wrong because you are forgetting that once the first card is picked out, it is not replaced. So the first card can be picked in 6 ways, but the second card can be picked in only 5 ways. So there are 30 outcomes.

loseyourname, not only does your approach assume that order does not matter (in one part and not the other), it more importantly neglects non-replacement.

 

[Didd]

Gokul43201 , you are right! What am I thinking.

I was tricked by my silly reasoning. Thank you for catching that.

 

[loseyourname]

This assumes order doesn't matter.

Well, it doesn't have to for the calculation to work.

If order didn't matter, the list would have 2 elements, not four.

AKG's analysis is correct.

That is, if I am consistent with the approach. Crap.