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The probability that a convenience store has cans of iced

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The probability that a convenience store has cans of iced [#permalink]

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New post 14 Jun 2010, 09:17
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The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that he will not be able to buy a can of iced tea?

A. \(\frac{1}{8}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{4}\)
E. \(\frac{7}{8}\)
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Apr 2012, 23:42, edited 2 times in total.
Edited the question, added the answer choices and OA

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Re: Probability [#permalink]

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New post 14 Jun 2010, 10:06
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The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that he will not be able to buy a can of iced tea?


Let the stores be A, B and C. The probability that each store does NOT have iced tea is 1/2

Therefore the probability that all 3 stores do not have ice tea in stock = 1/2 * 1/2 * 1/2

Ans: 1/8

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Re: Probability [#permalink]

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New post 29 Apr 2012, 22:04
Bunuel, can you please show how to solve this one?

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Re: Probability [#permalink]

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New post 29 Apr 2012, 23:41
Smita04 wrote:
Bunuel, can you please show how to solve this one?


The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that he will not be able to buy a can of iced tea?
A. \(\frac{1}{8}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{4}\)
E. \(\frac{7}{8}\)

James won't be able to buy a can of iced tea if none of the shops has it: P(NNN)=1/2^3=1/8.

Answer: A.

For other versions of this question check this: the-probability-that-a-convenience-store-has-cans-of-iced-128689.html
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Re: Probability [#permalink]

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New post 14 Feb 2013, 08:32
Bunuel wrote:
Smita04 wrote:
Bunuel, can you please show how to solve this one?


The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that he will not be able to buy a can of iced tea?
A. \(\frac{1}{8}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{4}\)
E. \(\frac{7}{8}\)

James won't be able to buy a can of iced tea if none of the shops has it: P(NNN)=1/2^3=1/8.

Answer: A.

For other versions of this question check this: the-probability-that-a-convenience-store-has-cans-of-iced-128689.html


bro bunuel,

how to decide if we have to add or if we have to multiply probabilities?
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Re: Probability [#permalink]

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New post 14 Feb 2013, 10:27
The probability that he won't be able to buy ice tea = p(NNN)=
1/2 *1/2 * 1/2 = 1/8

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Re: Probability [#permalink]

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New post 14 Feb 2013, 11:51
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Expert's post
Sachin9 wrote:
Bunuel wrote:
Smita04 wrote:
Bunuel, can you please show how to solve this one?


The probability that a convenience store has cans of iced tea in stock is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that he will not be able to buy a can of iced tea?
A. \(\frac{1}{8}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{4}\)
E. \(\frac{7}{8}\)

James won't be able to buy a can of iced tea if none of the shops has it: P(NNN)=1/2^3=1/8.

Answer: A.

For other versions of this question check this: the-probability-that-a-convenience-store-has-cans-of-iced-128689.html


bro bunuel,

how to decide if we have to add or if we have to multiply probabilities?


FUNDAMENTAL PRINCIPAL OF MULTIPLICATION :- if their are two jobs such that 1st job can be completed in 'm' ways and 2nd job can be completed in 'n' ways then two job successively can be done in m X n ways. i.e. doing 1st AND 2nd Job.
so in our case James didn't get the iced tea in 1st shop AND in 2nd shop AND in 3rd shop. -------- > (1/2) X (1/2) X (1/2) --------> (1/2)^3 ------> 1/8

Take another example :- In a class of 8 boys and 10 girls, teacher wants to select 1 boy AND 1 girl
number of ways selecting a boy = 8 ---------> number of ways selecting a girl = 10
so number of ways of selecting 1 boy AND 1 girl = 8 X 10 = 80

FUNDAMENTAL PRINCIPAL OF ADDITION :- if their are two jobs such that 1st job can be completed in 'm' ways and 2nd job can be completed in 'n' ways then either of the job can be performed in m + n ways. i.e. doing 1st OR 2nd Job.

Take the same example :- In a class of 8 boys and 10 girls, teacher wants to select 1 boy OR 1 girl (or we can say wants to select a student)
number of ways selecting a boy = 8 ---------> number of ways selecting a girl = 10
so number of ways of selecting 1 boy OR 1 girl = 8 + 10 = 18

Now in our case if the question had been what is the probability that james will get the iced tea in exactly one shop

he gets tea at 1st shop----------AND---didn't get at 2nd shop---AND---didn't get at 3rd shop------------(1/2) X(1/2) X (1/2)

---------------------------------------------------OR--------------------------------------------------------------------------------+

he didn't get tea at 1st shop---AND---gets at 2nd shop----------AND---didn't get at 3rd shop------------(1/2) X(1/2) X (1/2)

---------------------------------------------------OR--------------------------------------------------------------------------------+

he didn't get tea at 1st shop---AND---didn't get at 2nd shop---AND---gets at 3rd shop-------------------(1/2) X(1/2) X (1/2)

(1/8) + (1+8) + (1/8) = 3/8

After practice you will be able to solve it in a way (1/8)(3!/2!)

Hope it clears!

Regards,

Abhijit
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Re: Probability   [#permalink] 14 Feb 2013, 11:51
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