Does probability have a specific order in determining outcomes?

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The discussion centers on calculating the probability of selecting one smoking and one non-smoking table from a restaurant with equal numbers of each, chosen without replacement. The correct approach involves using conditional probabilities, where the probability of the second selection depends on the first. Participants clarify that the phrase "probability has no order" may refer to the idea that the sequence of events does not affect the overall probability when considering combinations. The calculations presented are validated, confirming the method used is appropriate for the scenario. Overall, the conversation emphasizes understanding the nuances of probability in relation to event ordering and selection.
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Homework Statement



I don't know how to ask my question without an example, so I am giving an example here.

For example, if there are 100 tables in a restaurant where 50 of them are for smoking while 50 of them are for non smoking. If two tables are chosen successively without replacement, then what is the probability of getting one tables for smoking and one for non-smoking?


Homework Equations





The Attempt at a Solution



P(required) = P(1st smoking)xP(2nd non smoking) + P(1st non smoking)xP(2nd smoking)

= 50/100 x 50/99 + 50/100 x 50/99


Is it correct? I am not sure. I remember some boooks saying that "probability has no order", I don't understand what it means actually.
 
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hi kenny1999! :smile:
kenny1999 said:
P(required) = P(1st smoking)xP(2nd non smoking) + P(1st non smoking)xP(2nd smoking)

= 50/100 x 50/99 + 50/100 x 50/99


Is it correct?

yes! :smile:

(what is worrying you about that? :confused:)
I remember some boooks saying that "probability has no order"

i've never seen that :confused:

what books?​
 
You wrote your symbolic answer slightly incorrectly, though the numerical work you did is as if you wrote the correct symbolic work. In other words, you knew what you meant, but you didn't write it right.

What you actually did was
P(1st smoking)*P(2nd non-smoking GIVEN 1ST SMOKING) + P(1st non-smoking)*P(2nd smoking GIVEN 1ST NON-SMOKEN)

This is how it works. Let A and B be two events.
P(A|B) = \frac{P(AB)}{P(B)} \rightarrow P(AB) = P(A|B)P(B)
P(B|A) = \frac{P(AB)}{P(A)}\rightarrow P(AB) = P(B|A)P(A)

So with other symbols, that string of probabilities is actually computing
P(1st smoking & 2nd non-smoking) + P(1st non-smoking & 2nd smoking)

And these are the only two ways to have 2 tables selected with differing types. So it is the probability of interest.
 
kenny1999 said:

Homework Statement



I don't know how to ask my question without an example, so I am giving an example here.

For example, if there are 100 tables in a restaurant where 50 of them are for smoking while 50 of them are for non smoking. If two tables are chosen successively without replacement, then what is the probability of getting one tables for smoking and one for non-smoking?


Homework Equations





The Attempt at a Solution



P(required) = P(1st smoking)xP(2nd non smoking) + P(1st non smoking)xP(2nd smoking)

= 50/100 x 50/99 + 50/100 x 50/99


Is it correct? I am not sure. I remember some boooks saying that "probability has no order", I don't understand what it means actually.

Your calculation is correct. Like Tiny-Tim, I, too have never heard the expression "probability has no order", although I do know that using probabilities to make ranking decisions can lead to nonsense (intransitivity). For example, if you say that A is better than B if P{A beats B} > 1/2, you can construct examples in which A is better than B, B is better than C and C is better than A. Maybe that is what your book meant.

RGV
 
kenny1999 said:

Homework Statement



I don't know how to ask my question without an example, so I am giving an example here.

For example, if there are 100 tables in a restaurant where 50 of them are for smoking while 50 of them are for non smoking. If two tables are chosen successively without replacement, then what is the probability of getting one tables for smoking and one for non-smoking?


Homework Equations





The Attempt at a Solution



P(required) = P(1st smoking)xP(2nd non smoking) + P(1st non smoking)xP(2nd smoking)

= 50/100 x 50/99 + 50/100 x 50/99


Is it correct? I am not sure. I remember some boooks saying that "probability has no order", I don't understand what it means actually.

I am not great at probability as I have just started learning about it, but does your book mean to convey the ideas of permutation and combination?

When your book says the probability has no order, does that mean the tables are round?
 
kenny1999 said:
I remember some boooks saying that "probability has no order", I don't understand what it means actually.

It would help if you could give some context for that statement or at least the names of the books. I suspect that you are misremembering.

I suspect you are you are thinking of a problem like "What is the probability of flipping two heads and three tails on five flips" as opposed to "What is the probability of flipping two heads and three tails on five flips in that order".
 
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