Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The probability that a convenience store has cans of iced [#permalink]
04 Mar 2011, 11:14

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

87% (01:11) correct
13% (02:22) wrong based on 15 sessions

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.

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 _________________

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?

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. _________________

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;

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

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

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?

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?

I´ve done an interview at Accepted.com quite a while ago and if any of you are interested, here is the link . I´m through my preparation of my second...

It’s here. Internship season. The key is on searching and applying for the jobs that you feel confident working on, not doing something out of pressure. Rotman has...