Finding probability based on a unknown value of another probability

[nicholaskong100]

Thought process: I know the sample space is 1. And I know I need to find the probability of the 2nd element because it shared among the probability of the 1st element and the 3rd element.

Most of the earlier problems I did gave a known value which made it easier to solve.

But how can I find the probability of the 2nd element IF I don't know what the probability of the 1st element or the 3rd element is?

 

[Dr.Peterson]

View attachment 28580

Thought process: I know the sample space is 1. And I know I need to find the probability of the 2nd element because it shared among the probability of the 1st element and the 3rd element.

Most of the earlier problems I did gave a known value which made it easier to solve.

But how can I find the probability of the 2nd element IF I don't know what the probability of the 1st element or the 3rd element is?

I don't know what your first line means; the sample space is S. Ah -- you meant the probability of S is 1. Correct.

And, yes, you're told how to find the other probabilities in terms of that of {2}.

If you let x = P({2}), can you express P({1}) and P({3}) in terms of x? Then you can solve an equation to find x.

 

[Harry_the_cat]

You have three equations with 3 unknowns. Solve simultaneously.

 

[nicholaskong100]

I don't know what your first line means; the sample space is S. Ah -- you meant the probability of S is 1. Correct.

And, yes, you're told how to find the other probabilities in terms of that of {2}.

If you let x = P({2}), can you express P({1}) and P({3}) in terms of x? Then you can solve an equation to find x.

Thanks. That's certainly makes thing easier to look at.

 

[JeffM]

Yes, well done.

You can do this two ways

P(1) = x, P(2) = y, and P(3) = z
x + y + z = 1
x = y + (1/6)
z = 2y

Or

(y + 1/6) + y + 2y = 1.

Either way leads yo the same place. Probabilities are simply real numbers in the interval [0, 1]. If they are mutually exclusive, you can add them.

 

[nicholaskong100]

Yes, well done.

You can do this two ways

P(1) = x, P(2) = y, and P(3) = z
x + y + z = 1
x = y + (1/6)
z = 2y

Or

(y + 1/6) + y + 2y = 1.

Either way leads yo the same place. Probabilities are simply real numbers in the interval [0, 1]. If they are mutually exclusive, you can add them.

Is mutually exclusive the same as disjoint meaning the subsets have outcomes that are different from the other subsets?

 

[Harry_the_cat]

Is mutually exclusive the same as disjoint meaning the subsets have outcomes that are different from the other subsets?

Yes.

 

[JeffM]

Is mutually exclusive the same as disjoint meaning the subsets have outcomes that are different from the other subsets?

Yes.

If two sets of outcomes have no members in common, the sets are disjoint or mutually exclusive. (I like to use the term "mutually exclusive" because it is simple English rather than math jargon.)

If sets A and B are mutually exclusive then

[math]\text {P(A and B)} = 0 \implies \text {P(A or B) = P(A) + P(B).}[/math]
It is very powerful. I think many of your questions over the last week or so revolved around this idea.