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Well, I've took around 1000hs in this...
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Well, I've took around 1000hs in this draw since I have the brokenst english you can imagine

Basically, there are those 9 balls, i want to know the probability of take the the 3 and the 6

They are introduced in a bag and you take two of them without see what are you taking, I suppose is the same take them one by one or take the two at the same time

Well, the draw says what i think is right, but I am not complety sure....anyway, in the worst case, it would be less than 2/9 right?

Can somebody help me with this?
>>
>>7811282

Well, I've taken around 1000 hours to complete this drawing because I have the most broken English you can imagine.

Basically, there are 9 balls. I want to know the probability of taking the 3-Ball and the 6-Ball.

They are mixed in the bag, and you can take two of them at random. It does not matter if you take them one by one or both at the same time.

Well the drawing says what I think is right, but I am not completely sure. Anyways, 2/9 would be the best probability you have, right?

Can somebody help me with this?
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>>7811282

yes your right. now get off from board before you introduce more broking english. and also do not say goodbye please or thank you. i think mine english got more brokenst just from reading OP
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>>7811300
Why are you so mean?

I only wanted to know about the ball's problem, i do not really care if my english is shit or if your english goes "brokenst", people can understand me, at least the first guy, who made fun of me

I thought this would be a fast thing to solve, you know, make a drawing, explain the problem, show a solution and wait for someone to tell me if I am right or not
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>>7811318

Sorry OP, I feel bad now so I'll give you a real answer:

Is it acceptable to choose them out of order (Is 6 then 3 the same as 3 then 6)?

After you pick the first ball, do you put it back in the bag?

---------------------- --------------------
NO then NO 1/72
NO then YES 1/81
YES then NO 2/72
YES then YES 2/81

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urn problem
$\displaystyle P=\frac{C(2,2)C(7,0)}{C(9,2)}= \frac{1\times 1}{36}=\frac{1}{36}$
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>>7811337
Thanks, so it's like I thought

But, is it the same if take the two balls at the same time? People normally do that, I think I would be the first girl to start to take them one by one
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>>7811282
if you pick two balls at random (from the bag), there's 72 possible outcomes.
Only (3,6) and (6,3) will lead to the event of interest therefore its probability is 2/72 = 1/36
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File: 1000hs'.jpg (54 KB, 900x875) Image search: [iqdb] [SauceNao] [Google]
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>>7811357
So, this is right too?

And is it the same if take the two balls at the same time? You know, only get my hand inside the bag once
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>>7811372
yes
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>>7811348