Why Do We Use the Ratio of Favorable to Possible Outcomes in Probability?

[Juwane]

Why do we take the "probability of an event" ratio as favorable / possible outcome

If A is an event, then probability that A will occur is given by

P(A)= no. of favorable outcomes / total no. of possible outcomes

Is this just a definition, or is there some special significance in taking the number of favorable outcomes as the numerator and total no. of possible outcomes as the denominator? Is it because since the numerator will always be less than the denominator, it will be easier to work with the ratio?

 

[farful]

Definitions.

Given a sample space, the probability of the entire sample space is 1, the probability of any element in the sample space is [0,1].

 

[Juwane]

Given a sample space, the probability of the entire sample space is 1, the probability of any element in the sample space is [0,1].

And how does this relate to the numerator being the number of favorable outcomes and the denominator being the total number of possible outcomes?

 

[SW VandeCarr]

And how does this relate to the numerator being the number of favorable outcomes and the denominator being the total number of possible outcomes?

"Favorable" is a judgment within some context. A probability simply relates the number of designated outcomes to all possible outcomes. The designated outcomes could be favorable, unfavorable or neutral depending on your point of view.

 

[Juwane]

My question is that why the favorable outcomes is in the numerator and total possible outcomes is in the denominator? Why not vice versa? Is it just a definition or does it have a significance?

 

[tiny-tim]

Hi Juwane!

Because for two independent events, you want both P(A + B) = P(A) + P(B), and P(AB) = P(A)P(B), and favourable/possible is the only way of doing that.

 

[Juwane]

I still don't understand. I like to understand through an example:

A coin is tossed. The probability that we will get a head is 1 over 2.

Another example:

A die is cast. The probability that we will get a 3 is 1 over 6.

Now my question is why, in the first example, it is 1 over 2 and not 2 over 1; and why in the second example it is 1 over 6 and not 6 over 1?

 

[tiny-tim]

… why in the second example it is 1 over 6 and not 6 over 1?

You could define it as 6 over 1, but then the probability of a 4 would be 6 over 1 also. and the probability of a 4 or a 3 would be 6 over 2, = 3 over 1.

And the probability of throwing the dice twice and getting a 4 first time and a 3 second time would be 36 over 1.

 

[Juwane]

You could define it as 6 over 1, but then the probability of a 4 would be 6 over 1 also. and the probability of a 4 or a 3 would be 6 over 2, = 3 over 1.

And the probability of throwing the dice twice and getting a 4 first time and a 3 second time would be 36 over 1.

So it's just the definition, right? Because it makes sense to say that the probability of getting a head in a coin throw is half, so it's 1/2, right?

 

[tiny-tim]

So it's just the definition, right? Because it makes sense to say that the probability of getting a head in a coin throw is half, so it's 1/2, right?

Right.

Mathematicians can define anything they like.

But a definition, even if perfectly valid, is only useful, if it successfully models the real world.

 

[g_edgar]

Well, you could take instead favorable outcomes / unfavorable outcomes and call it "odds".

So odds of 10 : 1 mean two probabilities of 10/11 and 1/11.

 

[robert Ihnot]

By Gee!, by Gosh! It was Laplace who DEFINED the probability of an event as the favorible outcomes divided by the total outcomes! It is the basic definition.

 

[bcramton]

Example: What is the probability that heads will fall on one throw of a coin? From theoretical probability, P(heads)=1/2. This means that 1/2 or 50% of your tosses of a coin should land on heads, theoretically. The numerator must be the favorable outcomes and the denominator must be the total possible outcomes. This ratio is 1 to 2 or 1:2, meaning, 1 out of every 2 tosses of the coin should result in heads. If you had possible outcomes in the numerator the ratio would be 2 : 1 , which would mean that you could toss 2 heads in 1 throw with one die. Impossible.

 

[Mensanator]

Example: What is the probability that heads will fall on one throw of a coin? From theoretical probability, P(heads)=1/2. This means that 1/2 or 50% of your tosses of a coin should land on heads, theoretically. The numerator must be the favorable outcomes and the denominator must be the total possible outcomes. This ratio is 1 to 2 or 1:2, meaning, 1 out of every 2 tosses of the coin should result in heads. If you had possible outcomes in the numerator the ratio would be 2 : 1 , which would mean that you could toss 2 heads in 1 throw with one die. Impossible.

2 possible outcomes doesn't mean they are both heads.

 

[DaveC426913]

A die is cast. The probability that we will get a 3 is 1 over 6.

Now my question is why... is it 1 over 6 and not 6 over 1?

1] What would it accomplish to have it defined as 6/1? Are there any ways this would be better?

2] There are a number of very useful reasons for defining it as 1/6. For example, you can add up all the possbilities and arrive at 1 again: 1/6+1/6+1/6+1/6+1/6+1/6 = 1. i.e. all individual possibilities add up to the total possible outcomes.

How would you do this if it were represented as 6/1?