dsstao
asked on
Multiple Probability
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!
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!
Gerwin Jansen
Hello, is this maybe homework?
ASKER
Haha, I'm 34 (and a college graduate without a math degree lol).
:)
20 bags, 10 marbles each of which one is blue - 200 marbles of which 20 are blue - chances are 20/200 - 10%
20 bags, 10 marbles each of which one is blue - 200 marbles of which 20 are blue - chances are 20/200 - 10%
ASKER
Hmm, but doesn't it matter that you're selecting a marble 20 different times?
"doesn't it matter that you're selecting a marble 20 different times?"
it sure does
it sure does
ASKER
aburr: Please elaborate... what does it do to the overall probability?
ASKER CERTIFIED SOLUTION
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Are you selecting 1 or 20 times (once in each bag)?
once in each bag
The 0.8 in the chance of failure in one bag
so the 1- failure is the overall chance of one sucess
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ASKER
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?
You are right. I should have used 0.9 instead of 0.8
"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
yes
even more if you use the correct 0.9 instead 0f the incorrect 0.8
ASKER
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?
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?
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.
ASKER
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.
"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..
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..
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.
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.
ASKER
Sorry aburr, I thought everyone was done commenting. Like I said, both sides make sense.
"Like I said, both sides make sense"
both sides may be reasonable but both sides cannot be right.
both sides may be reasonable but both sides cannot be right.
@dsstao - suggestion to open the question again? There is no definite answer yet...
ASKER
No, I don't think I will get a definitive answer unless a college math professor comes in and sets the record straight.
in that case, I think this question should be deleted.
ASKER
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.
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
ASKER
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.
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.
when events are independent, odds are multiplied. thus (0.9^20) chance of drawing a red ball 20 times