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M03-03

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M03-03  [#permalink]

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16 Sep 2014, 00:19
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76% (00:26) correct 24% (00:26) wrong based on 170 sessions

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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. $$\frac{7}{8}$$

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16 Sep 2014, 00:19
Official Solution:

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 stores has it. Since the probability that a convenience store has cans of iced tea is $$\frac{1}{2}$$ then the probability that a convenience store does not have cans of iced tea is also $$\frac{1}{2}$$, ($$1-\frac{1}{2}=\frac{1}{2}$$), so $$P(NNN)=( \frac{1}{2} )^3=\frac{1}{8}$$.

Answer: A
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Re: M03-03  [#permalink]

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07 May 2015, 03:29
Bunuel wrote:
Official Solution:

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 stores has it. Since the probability that a convenience store has cans of iced tea is $$\frac{1}{2}$$ then the probability that a convenience store does not have cans of iced tea is also $$\frac{1}{2}$$, ($$1-\frac{1}{2}=\frac{1}{2}$$), so $$P(NNN)=( \frac{1}{2} )^3=\frac{1}{8}$$.

Answer: A

WHAT IS WRONG WITH THE FOLLOWING APPROACH?

The probability that he will be able to buy a can of iced tea after visiting third store is 1/2*1/2*1/2=1/8.
so the probability that he wont be able to buy = 1-1/8=7/8?
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Posts: 54544
Re: M03-03  [#permalink]

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07 May 2015, 04:17
suhasreddy wrote:
Bunuel wrote:
Official Solution:

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 stores has it. Since the probability that a convenience store has cans of iced tea is $$\frac{1}{2}$$ then the probability that a convenience store does not have cans of iced tea is also $$\frac{1}{2}$$, ($$1-\frac{1}{2}=\frac{1}{2}$$), so $$P(NNN)=( \frac{1}{2} )^3=\frac{1}{8}$$.

Answer: A

WHAT IS WRONG WITH THE FOLLOWING APPROACH?

The probability that he will be able to buy a can of iced tea after visiting third store is 1/2*1/2*1/2=1/8.
so the probability that he wont be able to buy = 1-1/8=7/8?

P(BBB) = 1/8 is the probability that he buys in ALL 3 shops, so 1-1/8 is the probability that he will NOT be able to buy in ALL 3 shops, which is not not the same as not buying in any.
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Re: M03-03  [#permalink]

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10 May 2017, 01:36
Hi Bunuel.

Should we not consider that the person could visit the stores in any order and multiply the probability with 3! ?

Regards
Math Expert
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Re: M03-03  [#permalink]

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10 May 2017, 01:48
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dg88074 wrote:
Hi Bunuel.

Should we not consider that the person could visit the stores in any order and multiply the probability with 3! ?

Regards

No. We want P(No iced tea, No iced tea, No iced tea). NNN can be arranged only in one way. So, no need to multiply by 3!.
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Re: M03-03  [#permalink]

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10 May 2017, 02:22
Bunuel wrote:
dg88074 wrote:
Hi Bunuel.

Should we not consider that the person could visit the stores in any order and multiply the probability with 3! ?

Regards

No. We want P(No iced tea, No iced tea, No iced tea). NNN can be arranged only in one way. So, no need to multiply by 3!.

So out of curiosity, how should the question read if we had to multiply with 3! ?
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Joined: 09 Jul 2016
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GMAT 1: 730 Q50 V39
M03-03  [#permalink]

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Updated on: 27 May 2017, 10:00
Out of curiosity -

Is this approach correct ? -

Probability of him not able to buy after 1st store ( Failure in store 1) = 1/2
Probability of him not able to buy after 2nd store (Failure in store 1 x Failure in store 2) = 1/2 x 1/2
Probability of him not able to buy after 3rd store (Failure in store 1 x Failure in store 2 x Failure in store 3) = 1/2 x 1/2 x 1/2

Probability of him not getting in 3 store = Failure of above three probabilities = 1/2 + 1/4 + 1/8 = 7/8

@Bunnel - Can you please confirm whats wrong with above approach?

Originally posted by Vikram_Katti on 27 May 2017, 07:51.
Last edited by Vikram_Katti on 27 May 2017, 10:00, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 54544
Re: M03-03  [#permalink]

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27 May 2017, 08:55
Vikram_Katti wrote:
Out of curiosity -

Is this approach correct ? -

Probability of him not able to buy after 1st store ( Failure in store 1) = 1/2
Probability of him not able to buy after 2nd store (Failure in store 1 x Failure in store 2) = 1/2 x 1/2
Probability of him not able to buy after 3rd store (Failure in store 1 x Failure in store 2 x Failure in store 3) = 1/2 x 1/2 x 1/2

Probability of him not getting in 3 store = Failure of above three probabilities = 1/2 + 1/4 + 1/8 = 7/8

Can anyone confirm whats wrong with above approach?

Each successive case after the first one includes the previous one.

1/8 is already the probability that there is no candy in either of the stores.
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Joined: 09 Jul 2016
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GMAT 1: 730 Q50 V39
Re: M03-03  [#permalink]

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27 May 2017, 10:02
Bunuel wrote:
Vikram_Katti wrote:
Out of curiosity -

Is this approach correct ? -

Probability of him not able to buy after 1st store ( Failure in store 1) = 1/2
Probability of him not able to buy after 2nd store (Failure in store 1 x Failure in store 2) = 1/2 x 1/2
Probability of him not able to buy after 3rd store (Failure in store 1 x Failure in store 2 x Failure in store 3) = 1/2 x 1/2 x 1/2

Probability of him not getting in 3 store = Failure of above three probabilities = 1/2 + 1/4 + 1/8 = 7/8

Can anyone confirm whats wrong with above approach?

Each successive case after the first one includes the previous one.

1/8 is already the probability that there is no candy in either of the stores.

Oh! after re-reading it I see where I went wrong! Thanks Bunnel!
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Re: M03-03  [#permalink]

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30 Jun 2017, 23:16
Again one of the AND problem.

AND means multiplication.

He goes to first store "AND" he goes to second store "AND" he goes to third store.

1/2 * 1/2 * 1/2 = 1/8
Re: M03-03   [#permalink] 30 Jun 2017, 23:16
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