Stuck on probability measure

[nicholaskong100]

Solution:

 

[pka]

We know that in general we have [imath]\mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}(B)-\mathcal{P}(A\cap B)[/imath]
So [imath]1=\mathcal{P}(S)=\mathcal{P}(\{1,2\})+\mathcal{P}(\{2,3\})-\mathcal{P}(\{2\})=\dfrac{1}{3}+\dfrac{2}{3}-\mathcal{P}(\{2\})[/imath]
That means that [imath]\mathcal{P}(\{2\})=0[/imath] how & why?
Thus [imath]\mathcal{P}(\{1\})=?~\&~\mathcal{P}(\{3\})=?[/imath]

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[nicholaskong100]

Well, you add the fractions which gives you 3/3. 1 can also be written as 3/3. Subtract 3/3 with 3/3 and you get zero for the second sample point. P{1} = \frac{1}{3} and P{3} = \frac{2}{3}.

From what I am understanding if the problems have some givens, does that mean I can add all the givens = 1 as the sample space as how you wrote it?

 

[pka]

Well, you add the fractions which gives you 3/3. 1 can also be written as 3/3. Subtract 3/3 with 3/3 and you get zero for the second sample point. P{1} = \frac{1}{3} and P{3} = \frac{2}{3}.

From what I am understanding if the problems have some givens, does that mean I can add all the givens = 1 as the sample space as how you wrote it?

Well, using standard standard notation means the [imath]S=\{1,2,3\}[/imath] is the whole space.
So that [imath]\mathcal{P}(S)=1[/imath] What did you find for [imath]\mathcal{P}(\{1\})~\&~\mathcal{P}(\{3\})~?[/imath]

 

[Dr.Peterson]

View attachment 28460

Solution: View attachment 28461

Here is how I would start:

If P({1}) = x, P({2}) = y, and P({3}) = z, we have

P({1,2}) = 1/3 so x + y = 1/3​

P({2,3}) = 2/3, so y + z = 2/3​

P({1,2,3}) = 1, so x + y + z = 1​

Solve that system.

 

[nicholaskong100]

Here is how I would start:

If P({1}) = x, P({2}) = y, and P({3}) = z, we have

P({1,2}) = 1/3 so x + y = 1/3​

P({2,3}) = 2/3, so y + z = 2/3​

P({1,2,3}) = 1, so x + y + z = 1​

Solve that system.

Wow, that certainly makes thing much easier to look at. x = 1/3, y = 0 and z = 2/3.

 

[nicholaskong100]

Well, using standard standard notation means the [imath]S=\{1,2,3\}[/imath] is the whole space.
So that [imath]\mathcal{P}(S)=1[/imath] What did you find for [imath]\mathcal{P}(\{1\})~\&~\mathcal{P}(\{3\})~?[/imath]

x = 1/3, y = 0 and z = 2/3.