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The probability that a convenience store has no iced tea is

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The probability that a convenience store has no iced tea is [#permalink] New post 29 Nov 2007, 13:37
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The probability that a convenience store has no iced tea is 50%. If Karl is stopping by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?
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 [#permalink] New post 29 Nov 2007, 13:50
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7/8.

p=1-1/2^3(all stores have iced tea)=7/8
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Re: probability - at least one [#permalink] New post 11 May 2009, 15:22
Or...

p=p(the three stores don't have IT)+p(two stores don't have IT)+p(one store doesn't has IT)
p=1/2^3+1/2^2+1/2=7/8
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Re: probability - at least one [#permalink] New post 25 Jul 2009, 10:32
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atleast one ie => 1-(0.5)^3 => 0.875 => 7/8
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Re: probability - at least one [#permalink] New post 27 Sep 2009, 01:03
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Probability that atleast one store does not have
= 1 - Probability that (all stores have)
= 1 - (1/2) * (1/2) * (1/2)
= 1-(1/8)
= 7/8
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Re: probability - at least one [#permalink] New post 31 Jan 2010, 13:53
bmwhype2 wrote:
The probability that a convenience store has no iced tea is 50%. If Karl is stopping by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?


Even if question is: atleast one of the stores will have a can of iced tea
then the answer remains same. Isn't it?

Can anybody tell that?
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Re: probability - at least one [#permalink] New post 31 Jan 2010, 14:05
Expert's post
indir0ver wrote:
bmwhype2 wrote:
The probability that a convenience store has no iced tea is 50%. If Karl is stopping by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?


Even if question is: atleast one of the stores will have a can of iced tea
then the answer remains same. Isn't it?

Can anybody tell that?


Yes it is. 1-1/2^3=7/8. As the probabilities of having and not having the iced tea are 1/2, any "mirror" cases will have the same probabilities.
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Re: probability - at least one [#permalink] New post 14 Feb 2010, 08:16
bmwhype2 wrote:
The probability that a convenience store has no iced tea is 50%. If Karl is stopping by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?



probability that at least one of the stores will not have a can of iced tea = 1 - probability that all the stores will have a can of iced tea = 1 - (1/2 x 1/2 x 1/2) = 7/8
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Re: probability - at least one [#permalink] New post 08 Sep 2010, 02:07
My first response to this was 1/2*1/2*1/2 = 1/8

Since the probability for having iced tea/not having iced tea is 1/8 then why cant the above be true?
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Re: probability - at least one [#permalink] New post 25 Nov 2010, 21:43
at least one of the stores will not have a can of Ice tea = None of the stores will NOT have have a can of Ice tea = all the stores will have a can of Ice tea

50% probability a store does not have a can of Ice tea
therefores 50 % probability that a store will have a can of ice tea

1/2*1/2*1/2 = 1/8 (all have ice tea; 50% = .5 = 1/2)
1-1/8 = 7/8 (at least one store willl not have a can of ice tea)
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Re: probability - at least one [#permalink] New post 02 May 2011, 06:12
Hi,

P(atleast one does not have)= P(one does not have) + P( two don't have) + P(all 3 don't have)

= P(one does not have)*P(2 have) +
P (2 don't have)*P(one has) +
P(all 3 don't have)
= 1/2*1/2*1/2 + 1/4*1/2 + 1/8
= 3/8

Why is it giving me a wrong answer?I know I am missing something. :?

Can someone please explain?

Thanks..
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Re: probability - at least one [#permalink] New post 17 Jun 2011, 18:48
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at least 1 doesn't have = 1 doesnt have + 2 don't have + all 3 don't have
= 1/2 + 1/2*1/2 + 1/2*1/2*1/2 = 7/8
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Re: probability - at least one   [#permalink] 17 Jun 2011, 18:48
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