# 'And' means 'Multiply' and 'Or' means 'Add'?

1. Sep 28, 2015

### ajay.05

Mentor note: Moved from a non-HW forum section, so missing the HW template
When I was attending probability class, my math teacher said me that in the problems involving Case 1 'and' Case 2, I have to multiply probability of Case 1 and Case 2, while in the case of Case 1 'or' Case 2, I have to add probability of Case 1 and Case 2. I didn't get any valid reason from my Math teacher for this statement. Recently, when I was Studying Combinatorics, I cam across the same thing. I can give an Example from that...
"What is the number of ways of choosing 4 cards from a pack of playing cards, such that
1)Four Cards of the same suit
2)Four Cards belonging to different suits"

For the first question, it means 13C4 or 13C4 or 13C4 or 13C4, that is 13C4+13C4+13C4+13C4
For the second question, it means 13C1 and 13C1 and 13C1 and 13C1, that is 13C1*13C1*13C1*13C1
Can anybody please explain this to me? What do actually these and/or things mean?

Last edited by a moderator: Sep 28, 2015
2. Sep 28, 2015

### Michael27

If you look at the problem you have to figure out what it means to pick four cards from the deck in order to fulfill the requirement of either case. It is impossible to draw both situations at once so the two should be separate draws. Unless you keep on drawing after the first pick of four.
Your example does not tell me what to do.
If I would guess you mean that for case 1) you end up with a combination problem that doesn't involve random drawing just calculating the number of sequences that fit case 1) which is 52*12*11*10 combinations. No probability involved, just a number of sequences that are not unique. If you need them to be unique you need to figure out how many combination ABCD you need to eliminate from the combinations you already have. And remember you have four suits.
Your second case is very similar to the first but involves 52*13*13*13 combinations again these are not unique in a sense that a hand abcd is the same as abdc where a,b,c,d are cards in the deck.
So how does this relate to adding or multiplying is how you split up the problem. If you calculate combinations that can happen separate from each draw and have no overlap you can add all those combination together. If do not split up the problem you usually end up with multiplying each situation to get all the combinations. (and dividing out those that do not match yours)
I think you are making it hard because you mix up the terms probability and combinations. In order to calculate the probability of a person drawing four cards out of a random 52 card deck involves the number of combinations which are correct dividing them by the number that you actually can be drawn namely 52*51*50*49 if you are only interested in the first four draws and then resetting the deck.
I hope this helps.

3. Sep 29, 2015

### ajay.05

Yes! I do know that this involves only combinatorics...But I wondered, whether this principle is followed in combinatorics too?

4. Sep 29, 2015

### Michael27

In a sense it does.
Let {a,b,c,d} be an item in the result set S where a,b,c and d are cards the actions on those combinations can be seen as items in a set. When making an intersection of all the subsets you'll end up with satisfying your S1 and S2 and S3 and S4 question which is not exactly multiplication. When you do an Or operation you do a join of all your result sets which can be seen as an addition.
The Or and And should be seen as operations on your (sub)result set(s) and not as multiplications or additions.
Hope this helps a bit more.

5. Sep 30, 2015

### haruspex

It is important to understand that neither of those statements are true in general. Each only applies under specific (and different) conditions.

You can multiply probabilities to get the 'joint' probability (i.e. the probability that both events happen) only if the events are independent. That is, if knowing that one event occurs does not alter the probability that the other occurs.
For a simple example, consider tossing a coin twice. What is the probability of getting two heads? Since the outcome of the second toss is independent of the outcome of the first toss, the probability of getting a head on the first AND a head on the second is 1/2 * 1/2 = 1/4.
For the combinatorics equivalent, count the number of possible outcomes (HH, HT, TH, TT). Only one of the four is HH.
Now consider two events that are not independent. Throw a die once. The probability that the result is an even number is 1/2. The probability that it is greater than 3 is 1/2. But the probability that it is even AND greater than 3 is 1/3, not 1/4. (4 and 6 are each even and greater than 3.)

To be able to add probabilities to get the probability that either one or the other event happens, you need to know that the events are mutually exclusive, i.e. that they cannot both happen. (So, in particular, they are definitely not independent.)
The probability of either a 4 or a 6 from one roll of a die is 1/6 + 1/6 = 1/3, because they cannot both happen. But if you ask for the probability of either a 4 on the the first roll or a 6 on the second roll it is different, because they could both happen. Can you work out what this probability is?

Edited as per Insightful's insight in post 6.

Last edited: Oct 1, 2015
6. Sep 30, 2015

### insightful

This doesn't seem right.

7. Oct 1, 2015

### haruspex

Good catch. I meant of, of course, 1/6+1/6=1/3.

8. Oct 1, 2015

Thank You...