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January 31st, 2010, 12:38 AM   #1
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Probability of having one of two unknown bags

Bag #1
900 red eggs
100 blue eggs

Bag #2
100 red eggs
900 blue eggs

You are given one of these two bags (50/50 chance) randomly.

You perform a set of trials where you take out one egg randomly from your bag, note the color, then put it back in, and repeat.

You drew 8 red eggs and 2 blue eggs from your bag during this trial. What is the probabilty (%) that you have Bag #1?




I thought it was 50% because you could have drawn 8 red and 2 blue from bag #2 as well, since drawing eggs and putting them back does not change the bag's contents...
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January 31st, 2010, 10:33 AM   #2
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Re: Probability of having one of two unknown bags

Hello, daigo!

I solved this at another site . . . but here it is again.


Quote:


You are given one of these two bags randomly (50/50 chance).

You perform a set of trials where you take out one egg randomly from your bag,
[color=beige]. . [/color]note the color, then put it back in, and repeat.

You drew 8 red eggs and 2 blue eggs from your bag during this trial.
What is the probabilty (%) that you have Bag #1?

This is Conditional Probability.

[color=beige] .[/color][color=blue][1][/color]




[color=beige]. . [/color]

[color=beige]. . [/color]

[color=beige]. . [/color][color=beige] .[/color][color=blue][2][/color]




[color=beige]. . [/color]

[color=beige]. . [/color]

[color=beige]. . [/color][color=beige] .[/color][color=red](A)[/color]

[color=beige] .[/color][color=blue][3][/color]


Substitute [color=blue][2][/color] and [color=blue][3][/color] into [color=blue][1][/color]:

[color=beige]. . [/color]





~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~



This result make sense when we consider [color=red](A)[/color].

[color=beige]. . [/color]


If we have Bag 2, it is virtually impossible to draw 8 Reds and 2 Blues.

Thereofore, they must have come from Bag 1.

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January 31st, 2010, 12:11 PM   #3
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Re: Probability of having one of two unknown bags

Thanks. I was shown another method where I do 100 - ([1^8 * 100] / [9^8]) but I did not understand why that was the answer. Your solution clarified it better.
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