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Wine bottles

by umaa » Thu Jul 17, 2008 1:37 am
I have 2 different wine bottles. If I want to drink one glass of wine every day, each bottle contains enough wine for 2 glasses, and I select one bottle of wine randomly each day, what is the probability that I finish one bottle completely before opening the second one?

a. 1/4
b. 5/6
c. 1/6
d. 1/3
e. 2/3

answer is, d
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by pepeprepa » Thu Jul 17, 2008 3:47 am
You have 2 bottles: A and B
You have 6 possible combinations of choosing A or B during the 4 days
and 2 good results (AABB and BBAA), you drink in the same one bottle the first two days

--> 2/6=1/3 -- >d
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by pepeprepa » Thu Jul 17, 2008 3:47 am
You have 2 bottles: A and B
You have 6 possible combinations of choosing A or B during the 4 days
and 2 good results (AABB and BBAA), you drink in the same one bottle the first two days

--> 2/6=1/3 -- >d
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by louvre » Thu Jul 17, 2008 4:55 am
I have tough time understanding this. Here is my logic!

There are 2 bottles (NOT glasses), the question is the probability of selecting the same bottle in first 2 days i.e day1 and day2.

So it's the probability of selecting a(any) bottle on the day 1 AND selecting the same bottle on the day 2.

P(a/any) = 1
p(the same next day) = 1/2.

Isn't it 1*1/2 = 1/2

If I look at in terms of glasses, I can pass through it, but only stuck when using the bottles.

Thanks for your clarifications.
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by pepeprepa » Thu Jul 17, 2008 6:55 am
I write all the possible combinations. You can draw a "tree" (don't know if that is the right term) to see all possiilities:
AABB
BBAA
ABAB
BABA
ABBA
BAAB

For your probabilistic logic I don't catch why we do not find 1/3, can't help you sorry.
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by pepeprepa » Thu Jul 17, 2008 6:56 am
I write all the possible combinations. You can draw a "tree" (don't know if that is the right term) to see all possiilities:
AABB
BBAA
ABAB
BABA
ABBA
BAAB

For your probabilistic logic I don't catch why we do not find 1/3, can't help you sorry.
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by anksanks » Thu Jul 17, 2008 7:50 am
let the bottle be labelled a and b
and glasses be labeled a1, a2 for bottle a and b1 and b2 for bottle b

on first day,
probability for a1 = 1/4
probability for a2 = 1/4
adding up for a = 1/4 + 1/4 = 1/2

on day two,
assuming a1 was selected on day 1,
probability for a2 = 1/3
assuming a2 was selected on day 1,
probability for a1 = 1/3
total probability for a = 1/3 + 1/3 = 2/3

for two days, probability for a will be
1/2 * 2/3 = 1/3

Please correct me if I am wrong...
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by pepeprepa » Thu Jul 17, 2008 8:34 am
anksanks, what i don't catch is why the probability of getting a2, assuming you got a1 the first day, is 1/3
I mean you do not choose a2 among 3 glasses of wine (given a1 is finished) but you choose between a bottle and another one, so 2 choices.
Tell me if I am wrong, I had a long day
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by anksanks » Thu Jul 17, 2008 8:51 am
What i understood from the problem is it is taking individual glasses and not bottles into consideration.
like there are two red balls and two green balls and not red color or green color.

The answers will totaly change if we assume it that way.

My first understanding was to divide it into four individual glasses like we divide it in case of colored balls or other probability issues.

I hope it helps.
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