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January 31st, 2010, 12:38 AM  #1 
Senior Member Joined: Jan 2010 Posts: 205 Thanks: 0  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... 
January 31st, 2010, 10:33 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Probability of having one of two unknown bags Hello, daigo! I solved this at another site . . . but here it is again. Quote:
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.  
January 31st, 2010, 12:11 PM  #3 
Senior Member Joined: Jan 2010 Posts: 205 Thanks: 0  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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