#1
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Interesting Conditional Probability Question
You have three bags, each containing 2 coins. The first bag has two gold coins, the second bag has one gold coin and one silver coin, and the third bag has two silver coins.
A bag is picked at random and a coin is picked at random, the coin is golden. The gold coin is returned to the bag which is shaken, and the bag is drawn from again. Again, the coin is gold. This coin is returned, the bag is shaken, and you draw a coin from the bag for a third time. What is the probability that this coin is golden? |
#2
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Re: Interesting Conditional Probability Question
Well, you don't have the silver/silver bag.
So you either have the gold/gold (GG) bag or the gold/silver (GS) bag. Pulling gold twice from the GS bag will happen 1 in 4 times, and of course 1 in 1 times for the GG bag. Hmmm... off the top of my head I con't figure out how to do this. My gut tells me you have an 80% chance of having the GG bag, making the probability for the next pull to be golden 90%. Little help? |
#3
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Re: Interesting Conditional Probability Question
Wait, picking a random bag (3 possibilities) and making 2 pulls (2 x 2 possibilities) gives 12 possible outcomes, only 5 of which are pulling two gold coins. Four of five of those have you holding the GG bad and one has you holding the GS bag. I think my gut was right.
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#4
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Re: Interesting Conditional Probability Question
[ QUOTE ]
You have three bags, each containing 2 coins. The first bag has two gold coins, the second bag has one gold coin and one silver coin, and the third bag has two silver coins. A bag is picked at random and a coin is picked at random, the coin is golden. The gold coin is returned to the bag which is shaken, and the bag is drawn from again. Again, the coin is gold. This coin is returned, the bag is shaken, and you draw a coin from the bag for a third time. What is the probability that this coin is golden? [/ QUOTE ] Possibility 1: You picked the bag with both golds and then picked 2 golds. This had initial probability (1/3)*1 = 1/3. Possibility 2: You picked the bag with 1 gold and 1 silver and then picked 2 golds. This had inital probability (1/3)*(1/4) = 1/12. Possibility 1 is 4 times more likely than possibility 2, so it is a 4:1 favorite, so the probability is 4/5 that you have the bag with 2 golds, and 1/5 that you have the bag with 1 of each. If you have the bag with 2 golds, the probability that the next one is gold is 1, and if you have the bag with one of each, the probability that the next one is gold is 1/2, so the probability that the next one is gold is: (4/5)*1 + (1/5)*(1/2) = 9/10 or 90%. |
#5
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Re: Interesting Conditional Probability Question
[ QUOTE ]
Possibility 1: You picked the bag with both golds and then picked 2 golds. This had initial probability (1/3)*1 = 1/3. Possibility 2: You picked the bag with 1 gold and 1 silver and then picked 2 golds. This had inital probability (1/3)*(1/4) = 1/12. Possibility 1 is 4 times more likely than possibility 2, so it is a 4:1 favorite, so the probability is 4/5 that you have the bag with 2 golds, and 1/5 that you have the bag with 1 of each. If you have the bag with 2 golds, the probability that the next one is gold is 1, and if you have the bag with one of each, the probability that the next one is gold is 1/2, so the probability that the next one is gold is: (4/5)*1 + (1/5)*(1/2) = 9/10 or 90%. [/ QUOTE ] Woot! |
#6
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Re: Interesting Conditional Probability Question
[ QUOTE ]
Woot! [/ QUOTE ] Translation: now he doesn't have to do his homework himself! |
#7
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Re: Interesting Conditional Probability Question
Not my homework problem. I was wooting the fact that I aparently answered correctly.
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#8
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Re: Interesting Conditional Probability Question
[ QUOTE ]
[ QUOTE ] Woot! [/ QUOTE ] Translation: now he doesn't have to do his homework himself! [/ QUOTE ] SHIP IT! |
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