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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] New post 04 Mar 2011, 11:14
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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}


Guys, can someone explain me how is it possible that the chances for him to buy and not to buy is the same probability?

if he have 1/8 chance not to find it, how much chance he will have to find it and why. thanks.
[Reveal] Spoiler: OA

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Re: m03 #3 [#permalink] New post 04 Mar 2011, 11:26
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Since the probability of the event occurring is equally likely as to the probability of the event not occurring, then both the chances of finding the can or not finding the can after visiting 3 stores will be 1/8.

I see it like this;

If he goes to the first store and finds the can. He won't go to the next store, right!!

But, if the question says; he visited 3 stores for the can, we can infer that he didn't find it in the first two stores.

P(Finding) = 50% = 1/2

P(Not finding) = 1/2

He visited 3 stores; he didn't find it;

He didn't find in 1st store AND he didn't find it in 2nd store AND he didn't find it in 3rd store
1/2*1/2*1/2(this last 1/2 is the probability for Not finding the can even in the 3rd store) = 1/8

He visited 3 stores; he found it(in the 3rd store);
He didn't find in 1st store AND he didn't find it in 2nd store AND he FOUND it in 3rd store
1/2*1/2*1/2(this last 1/2 is the probability for successfully finding the can) = 1/8
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 11:28
so when do u use the 1-p? thanks +1
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 11:35
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144144 wrote:
hey all


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?

* \frac{1}{8}
* \frac{1}{4}
* \frac{1}{2}
* \frac{3}{4}
* \frac{7}{8}


Guys, can someone explain me how is it possible that the chances for him to buy and not to buy is the same probability?

if he have 1/8 chance not to find it, how much chance he will have to find it and why. thanks.


No, above is not correct. James will be able to buy a can of iced tea if at least one of the shops has it and the probability that at least one of the shops has an iced tea is 7/8 (so it is indeed 1-1/8=7/8).

Each shop has 2 options either to have an iced tea or not, so there are 2*2*2=8 outcomes possible. James will not be able to buy a can of iced tea if neither of the shops has it, P=1/2*1/2*1/2 in all other 7 options at least one has it thus he'll be able to buy it.
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 11:39
WOW amazing explanation. thanks as usual.
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 11:43
so the exact way to calculate will be (only for my understanding)

find on the 1st store - 1/2
2nd store - 1/2*1/2 = 1/4
3rd store - 1/2*1/2*1/2=1/8

so 1/2+1/4+1/8 = 7/8 for him to find it.

am i right?

thanks.
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 11:49
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144144 wrote:
so the exact way to calculate will be (only for my understanding)

find on the 1st store - 1/2
2nd store - 1/2*1/2 = 1/4
3rd store - 1/2*1/2*1/2=1/8

so 1/2+1/4+1/8 = 7/8 for him to find it.

am i right?

thanks.


Direct probability approach:
One store has an iced tea: P(YNN)=3*1/2*1/2^2=3/8 (multiplying by 3 as scenario YNN can occur in 3!/2!=3 ways);
Two shops have an iced tea: P(YYN)=3*1/2^2*1/2=3/8;
All three shops have an iced tea: P(YYY)=1/2^3=1/8;

P=3/8+3/8+1/8=7/8.
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 11:57
got it. thanks.
so i guess the way i did it was lucky and it is wrong to do so...
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 12:14
144144 wrote:
got it. thanks.
so i guess the way i did it was lucky and it is wrong to do so...


I dont think that the way you did it is wrong. Let me try and verbalise wha you tried to do mathematically.

The event here is being able to buy ice tea as one passes three stores. The total probability of this event will be equal to the sum of following three events:

1. Probability that it is available in first store (then whether it is available in second store and third store or not is immaterial) = 1/2

2. Probability that it is not available in first store but available in second (immaterial whether available in third or not), so 1/2*1/2 = 1/4

3. Probability that it is not available in first and second store but available in third = 1/2*1/2*1/2 = 1/8

Total probability = 1/2+1/4+1/8 = 7/8
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 17:16
this is the concept of atleast one i.e. 1- x.
x = no luck in any store

Btw OA for this is correct. Isnt it?

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?

beyondgmatscore wrote:
144144 wrote:
got it. thanks.
so i guess the way i did it was lucky and it is wrong to do so...


I dont think that the way you did it is wrong. Let me try and verbalise wha you tried to do mathematically.

The event here is being able to buy ice tea as one passes three stores. The total probability of this event will be equal to the sum of following three events:

1. Probability that it is available in first store (then whether it is available in second store and third store or not is immaterial) = 1/2

2. Probability that it is not available in first store but available in second (immaterial whether available in third or not), so 1/2*1/2 = 1/4

3. Probability that it is not available in first and second store but available in third = 1/2*1/2*1/2 = 1/8

Total probability = 1/2+1/4+1/8 = 7/8
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 17:33
This one from Gmatprep so I remember it. It is analogous to the above question. I think :wink:

On his drive to work, Leo listens to one of 3 radio stations, A, B, or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns to B. If B is playing a song that he likes, he listens to it; if not, he turns to C. If C is playing a song he likes, he listens; if not, he turns off the radio. For each station, the probability is 0.3 that at any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Leo ill hear a song he likes?

a. 0.027
b. 0.09
c. 0.417
d. 0.657
e. 0.9

[Reveal] Spoiler:
D
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Re: m03 #3 [#permalink] New post 04 Mar 2011, 22:37
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so
p(YYY) = 3/10*3/10*3/10 = 9/1000
P(YYN) = 3*3/10*3/10*7/10 = 189/1000
P(YNN) = 3*3/10*7/10*7/10 = 441/1000

its not the right answer... what am i missing? thanks
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Re: m03 #3 [#permalink] New post 05 Mar 2011, 00:12
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144144 wrote:
so
p(YYY) = 3/10*3/10*3/10 = 9/1000 is actually 27/1000
P(YYN) = 3*3/10*3/10*7/10 = 189/1000
P(YNN) = 3*3/10*7/10*7/10 = 441/1000

its not the right answer... what am i missing? thanks


Just a minor calculation mistake.
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Re: m03 #3 [#permalink] New post 05 Mar 2011, 00:34
GEEZ - Im an idiot. thanks Fluke.
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Re: m03 #3 [#permalink] New post 05 Mar 2011, 01:53
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gmat1220 wrote:
This one from Gmatprep so I remember it. It is analogous to the above question. I think :wink:

On his drive to work, Leo listens to one of 3 radio stations, A, B, or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns to B. If B is playing a song that he likes, he listens to it; if not, he turns to C. If C is playing a song he likes, he listens; if not, he turns off the radio. For each station, the probability is 0.3 that at any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Leo ill hear a song he likes?

a. 0.027
b. 0.09
c. 0.417
d. 0.657
e. 0.9

[Reveal] Spoiler:
D


Discussed here: radio-station-104659.html and can be solved as 144144 suggested in this post: m03-110379.html#p884977

The desired probability is the sum of the following events:

A is playing the song he likes - 0.3;
A is not, but B is - 0.7*0.3=0.21;
A is not, B is not, but C is - 0.7*0.7*0.3=0.147;
P=0.3+0.21+0.147=0.657.

Or: 1-the probability that neither of the stations is playing the song he likes: P=1-0.7*0.7*0.7=0.657.

Answer: D.
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Re: m03 #3   [#permalink] 05 Mar 2011, 01:53
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