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

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Difficulty:   5% (low)

Question Stats: 78% (00:40) correct 22% (01:00) 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. $$\frac{7}{8}$$

Originally posted by Praetorian on 14 Jun 2010, 08:17.
Last edited by Bunuel on 30 Sep 2019, 22:37, edited 3 times in total.
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Re: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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Quote:
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: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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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.
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Re: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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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: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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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: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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WOW amazing explanation. thanks as usual.
Senior Manager  Joined: 08 Nov 2010
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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: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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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: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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This one from Gmatprep so I remember it. It is analogous to the above question. I think 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

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Re: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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gmat1220 wrote:
This one from Gmatprep so I remember it. It is analogous to the above question. I think 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

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.

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

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

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.

For other versions of this question check this: http://gmatclub.com/forum/the-probabili ... 28689.html
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Re: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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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.

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: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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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.

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: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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Bruenell, I believe that the solution should be 1/2. The person is going to three stores, that means probability of not getting the can in the first two stores is 1 and the last one is 1/2. hence it should be 1*1*1/2
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Re: The probability that a convenience store has cans of iced tea in stock  [#permalink]

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Bruenell, I believe that the solution should be 1/2. The person is going to three stores, that means probability of not getting the can in the first two stores is 1 and the last one is 1/2. hence it should be 1*1*1/2

Why is the probability of not getting ice tea in the first two shops 1? The question clearly says that "The probability that a convenience store has cans of iced tea in stock is 1/2", so the probability of not having is also 1/2. P(No, No, No) = 1/2^3 = 1/8.
_________________ Re: The probability that a convenience store has cans of iced tea in stock   [#permalink] 01 Jun 2020, 23:40

# The probability that a convenience store has cans of iced tea in stock   