Basic Probability Theory (Equaly Likely Principle)

[FaraDazed]

Homework Statement

Five cards numbered 1 to 5 are shuffled and placed face down on a table. Two of the cards are picked at random. [Hint: find all of the possible outcomes of this experiment which form the sample space S and use the Equally Likely Principle.]

Find the probability of the following events:

A, the number 1 is selected;

B, the smaller of the two numbers selected is 2;

C, the difference of the numbers is even

Homework Equations

The probability of an event A is defined as the proportion of outcomes in the sample space S that correspond to A:
P(A)=n(A)/n(S)

The Attempt at a Solution

I am very new to this, I wasn't sure of something and needed it clearing up before I can give a definite answer for each part. Does anyone know if, in this problem, out of the two cards that are picked is picking a 1 and 2, the same as picking a 2 and 1 (if you know what I mean).

For part A then: On the basis of the equally likely principle, if they are considered the same then the total possible outcomes are 10 and then the answer would be 2/10=1/5. If they are not considered the same then the total possible outcomes becomes 20 and the answer would become 2/20=1/10.

For Part B I am not sure it matter if they are the same or not but from what I can gather, the total possible outcomes are 3 (2 and 3, 2 and 4, 2 and 5). so the answer would be 2/3.

For Part C if they are the same then the total possible outcomes is 4 so the answer would be 2/4=1/2. If they are not the same then the total possible outcomes would be 8 and therefore the answer would be 2/8=1/4.

The only downside to my "go" at this problem is that to find the possible number of outcomes for things I had to pen them all down and cross them off. If there was a better way to do that I would appreciate some advice on that :) .

Thanks :)

 

[PeroK]

Why not get yourself a pack of cards and experiment and see what happens?

 

[FaraDazed]

Why not get yourself a pack of cards and experiment and see what happens?

Well, you could pick the 2 up first and then the 1 up second and you could also pick the 1 up first and then 2 second. But once you have them in your hands, they are the same two cards regardless of order. That is what is confusing me.

 

[PeroK]

You can solve the problem either way, as long as you are consistent:

You can consider 20 (5x4) possibilities for 2 cards; of which 4 have 1 as the first card and 4 have 1 as the second card.

Or, you consider only 10 combinations; of which only 4 have a 1.

It's up to you; whatever makes the calculations simpler and more logical.

 

[FaraDazed]

You can solve the problem either way, as long as you are consistent:

You can consider 20 (5x4) possibilities for 2 cards; of which 4 have 1 as the first card and 4 have 1 as the second card.

Or, you consider only 10 combinations; of which only 4 have a 1.

It's up to you; whatever makes the calculations simpler and more logical.

Ah right ok, thanks. So I guess my answers are incorrect seeing as I got different answers depending on which way I looked at it.

 

[PeroK]

Doing a problem two different ways is a good way to confirm you've got the correct answer. But, of course, if you get different answers, then at least one of them must be wrong.

It's still a good idea to try it out with real cards and see what happens.

 

[Ray Vickson]

Well, you could pick the 2 up first and then the 1 up second and you could also pick the 1 up first and then 2 second. But once you have them in your hands, they are the same two cards regardless of order. That is what is confusing me.

Doing the problem two different ways is good, provided that you do both ways correctly. Physically, you really do pick one card first, then pick a second card afterwards. Therefore, in (A) the two outcomes (first,second) = (1, x) or (x,1) (where x = 2, 3, 4 or 5) describers exactly what constitutes the event "pick a 1". In (B) the event "smaller is 2" = (2,x) or (x,2), where x = 3, 4 or 5. In (C) the event "difference is even" = {(a,b): |b-a|=0 or |b-a| = 2 or |b-a| = 4}.