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Multiple Probability

Posted on 2012-04-06
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Last Modified: 2012-04-17
Hi All,

I have a hopefully simple question about probability:

If I have a bag of 10 marbles where 9 are red and 1 is blue, the probability of me blindly reaching into that bag and grabbing the blue one is 10%.

However, let's say I now have 20 bags of marbles, populated exactly as above.  What are my chances of grabbing a blue marble in any of the bags?

Thanks!
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Question by:dsstao
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31 Comments
 
LVL 38

Expert Comment

by:Gerwin Jansen, EE MVE
ID: 37817294
Hello, is this maybe homework?
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LVL 1

Author Comment

by:dsstao
ID: 37817320
Haha, I'm 34 (and a college graduate without a math degree lol).
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LVL 38

Expert Comment

by:Gerwin Jansen, EE MVE
ID: 37817360
:)

20 bags, 10 marbles each of which one is blue - 200 marbles of which 20 are blue - chances are 20/200 - 10%
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LVL 1

Author Comment

by:dsstao
ID: 37817397
Hmm, but doesn't it matter that you're selecting a marble 20 different times?
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LVL 27

Expert Comment

by:aburr
ID: 37817413
"doesn't it matter that you're selecting a marble 20 different times?"
it sure does
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LVL 1

Author Comment

by:dsstao
ID: 37817425
aburr: Please elaborate... what does it do to the overall probability?
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LVL 27

Accepted Solution

by:
aburr earned 1000 total points
ID: 37817428
1-0.8^20
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LVL 38

Expert Comment

by:Gerwin Jansen, EE MVE
ID: 37817430
Are you selecting 1 or 20 times (once in each bag)?
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LVL 27

Expert Comment

by:aburr
ID: 37817434
once in each bag
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LVL 27

Expert Comment

by:aburr
ID: 37817436
The 0.8 in the chance of failure in one bag
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LVL 27

Expert Comment

by:aburr
ID: 37817441
so the 1- failure is the overall chance of one sucess
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LVL 38

Assisted Solution

by:Gerwin Jansen, EE MVE
Gerwin Jansen, EE MVE earned 1000 total points
ID: 37817462
Where is the 0.8 from btw? 1 out of 10 leaves a 0.9 chance of failure.

So chance of getting all blue marbles in 20 bags is 0.1^20 or 0.0.00000000000000000001
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LVL 1

Author Comment

by:dsstao
ID: 37817465
1 - (0.8^20) = 0.988470785, so that means if I have those 20 bags, and select one marble in each bag, that I have a 98.8% chance of eventually getting a blue one?
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LVL 27

Expert Comment

by:aburr
ID: 37817490
You are right. I should have used 0.9 instead of 0.8
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LVL 27

Expert Comment

by:aburr
ID: 37817495
"1 - (0.8^20) = 0.988470785, so that means if I have those 20 bags, and select one marble in each bag, that I have a 98.8% chance of eventually getting a blue one?"
yes
even more if you use the correct 0.9 instead 0f the incorrect 0.8
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LVL 1

Author Comment

by:dsstao
ID: 37817514
1 - (0.9^20) = 0.878423345 or about 87.8%.

To Sum Up:
20 Bags
10 Marbles/Bag
1 Blue Marble/Bag

If I reach in to each of the 20 bags, my chance of coming up with at least 1 blue marble is 87.8%.

Sound right?
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LVL 83

Expert Comment

by:Dave Baldwin
ID: 37817553
No, no, no.  Each bag represents a separate event.  You have a 10% chance 20 times.  While you can accumulate statistics, the odds can't be combined because they are separate Unrelated events.
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LVL 1

Author Comment

by:dsstao
ID: 37817631
Lol, alright, I see a lot of answers from different people and no way to distinguish who's right.  No offense to anyone, but I can see how each answer might make sense.
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LVL 27

Expert Comment

by:aburr
ID: 37817813
"the odds can't be combined because they are separate Unrelated events."
But they are not separate events. The combined 20 draws (from different bags) is ONE defined event as defined by the author of this question..
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LVL 27

Expert Comment

by:aburr
ID: 37817817
I see that I am late in posting my comment. The answer is NOT 10%
That can easily be seen by looking at the first two draws. On the first draw you have a 10% chance of getting a blue ball. On the second draw you have an additional 10% chance. Clearly two 10% chances give you a better than 10% chance of getting a blue ball.
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LVL 1

Author Comment

by:dsstao
ID: 37818020
Sorry aburr, I thought everyone was done commenting.  Like I said, both sides make sense.
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LVL 27

Expert Comment

by:aburr
ID: 37818156
"Like I said, both sides make sense"
both sides may be reasonable but both sides cannot be right.
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LVL 38

Expert Comment

by:Gerwin Jansen, EE MVE
ID: 37818539
@dsstao - suggestion to open the question again? There is no definite answer yet...
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LVL 1

Author Comment

by:dsstao
ID: 37819040
No, I don't think I will get a definitive answer unless a college math professor comes in and sets the record straight.
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LVL 38

Expert Comment

by:Gerwin Jansen, EE MVE
ID: 37819449
in that case, I think this question should be deleted.
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LVL 1

Author Comment

by:dsstao
ID: 37819593
I'm not sure that I can delete a question with points already assigned.  Let's just let it go.  If you disagree, you can contact a Mod about it.
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LVL 38

Expert Comment

by:Gerwin Jansen, EE MVE
ID: 37819960
well, I'm no mod but you can, use request attention button on top. just that for future references, this question has not been answered. @Dave - no points on this one, sorry
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LVL 1

Author Comment

by:dsstao
ID: 37819967
I will judge whether the question has been answered or not, and in my view, "aburr" makes more sense than you do.  Your persistence in trying to get me to do something I don't want to do makes me lean towards you even less.

Oh, and he didn't waste my time asking if it was a homework assignment first.

This will be my last post or my last time ever reviewing this question.
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LVL 84

Expert Comment

by:ozo
ID: 37820497
when events are independent, odds are multiplied.  thus (0.9^20) chance of drawing a red ball 20 times
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