a problem with " law of total probability "

[acoustic fox]

here we have 2 conatainers of elements, let's say they're blue and red balls.

in the first we have 4 red, 2 blue and the total of 6 balls
in the second we have 3 red, 4 blue and the total of 7 balls

let's pick on ball out of these balls.
we want to calculate the probability of the blue ball ( how much chance does it have of being chosen )

why the answers are different when ( in the second method ) we assume that each conatiner is a partition of a sample sapce ( as shown in the diagram ).

( sorry the answer in the first method is 19/42 )

can someone tell me what am i doing wrong? why do i not get the same answer i did using the " tree " method, multiplying probabilities?

[shell mortar]

What is P(Blue)?

[acoustic fox]

probability of the blue

also sorry i made a mistake in the sum

[shell mortar]

What's the original question?

[acoustic fox]

uhm

ok i include it in the text sorry

i hope i have clarified the problem enough

[shell mortar]

Oh no that was clear but the question is the following:
Do you pick one ball from both the containers and want atleast one to be blue? Which I guess will have the equivalent as you pick two balls from the aggregate and one of them should be blue or is it something different

[acoustic fox]

no, we pick one ball out of the total

no matter whch container

1 ball out of total

[shell mortar]

So you mix the balls and then you pick one

[acoustic fox]

so... i think my problem lies here

does mixing make difference?

i assumed the 2 conatiners as 2 partitions

[shell mortar]

Yes

Mixing the balls and then picking one changes the outcome

[acoustic fox]

why? is this the reason we cannot represent the 2 containers using partitions in a singe sample space?

[shell mortar]

Well because your action changes which is why I was confused about the initial P(blue)

In the first scenario, where you have two boxes

You first pick a box at random

and then you pick a ball from that at random

Which is what you're calculating in the tree method

Note that once you have picked a box, the other box's balls don't affect the ball you chose from the said box

Whereas when you mix the two bags

you don't have to choose between them

and the total balls are what you choose from

Those are two different actions

[acoustic fox]

hmmmm

[acoustic fox]

about this

can't we say the same thing happens with partitions? each partition has no mutuality with the other, therefore can't we say " ... so no elements within one partition has no effects on the other elements within different partitions "?

[shell mortar]

If you can then the sample space of one partition will not include the balls from the other partition

[acoustic fox]

isn't this the definition a partition of a sample space?

subsets within a single sample space that have nothing in mutual

[acoustic fox]

can you please also tell me where i can read more about this " mixture " thing? i kinda find it hard to get, i'm not really good at this

[shell mortar]

Mixing would just be putting all the balls together in a single box with no partitions

[acoustic fox]

so this means i did not mix these balls? i drew a partition line in the diagram

[acoustic fox]

thank you very much

/close

close

.close

[brittle wasp]

Solved

Post marked as solved by @acoustic fox.

Use .unsolved if this was a mistake.

[acoustic fox]

solved

.colved