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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 06 Mar 2012, 12:26
1. 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 be able to buy a can of iced tea?

2. The probability that a convenience store has no iced tea is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?

3. 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?
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Re: probability [#permalink] New post 06 Mar 2012, 12:44
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1. 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 be able to buy a can of iced tea?

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 P=1-(none of the shops has an iced tea)=1-P(NNN)=1-1/2^3=7/8.

2. The probability that a convenience store has no iced tea is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?

The probability that at least one of the stores will not have a can of iced tea is 1-(all shops have an iced tea)=1-P(YYY)=1-1/2^2=7/8. Notice that we have the same answer as above. That's because the probability distribution is symmetrical for this question thus the probability that at least one of the stores will NOT have a can of iced tea = the probability that at least one of the stores will have a can of iced tea

3. 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?

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.

Hope it's clear.
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Re: probability [#permalink] New post 17 Nov 2012, 00:13
Bunuel wrote:
1. 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 be able to buy a can of iced tea?

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 P=1-(none of the shops has an iced tea)=1-P(NNN)=1-1/2^3=7/8.

2. The probability that a convenience store has no iced tea is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?

The probability that at least one of the stores will not have a can of iced tea is 1-(all shops have an iced tea)=1-P(YYY)=1-1/2^2=7/8. Notice that we have the same answer as above. That's because the probability distribution is symmetrical for this question thus the probability that at least one of the stores will NOT have a can of iced tea = the probability that at least one of the stores will have a can of iced tea

3. 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?

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.

Hope it's clear.



Hi Bunuel,

WRT Q2.


P(atleast one does not have)= P(one does not have) + P( two don't have) + P(all 3 don't have)

= P(one does not have)*P(2 have) +
P (2 don't have)*P(one has) +
P(all 3 don't have)
= 1/2*1/2*1/2 + 1/4*1/2 + 1/8
= 3/8

Why is it giving me a wrong answer?I know I am missing something.

Can someone please explain?

Infact Is my methos wrong, coz i am trying to solve Qs by this method. I understood the 1-p methos as well.
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Re: probability [#permalink] New post 17 Nov 2012, 05:53
Expert's post
aditi2013 wrote:
Bunuel wrote:
1. 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 be able to buy a can of iced tea?

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 P=1-(none of the shops has an iced tea)=1-P(NNN)=1-1/2^3=7/8.

2. The probability that a convenience store has no iced tea is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?

The probability that at least one of the stores will not have a can of iced tea is 1-(all shops have an iced tea)=1-P(YYY)=1-1/2^2=7/8. Notice that we have the same answer as above. That's because the probability distribution is symmetrical for this question thus the probability that at least one of the stores will NOT have a can of iced tea = the probability that at least one of the stores will have a can of iced tea

3. 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?

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.

Hope it's clear.



Hi Bunuel,

WRT Q2.


P(atleast one does not have)= P(one does not have) + P( two don't have) + P(all 3 don't have)

= P(one does not have)*P(2 have) +
P (2 don't have)*P(one has) +
P(all 3 don't have)
= 1/2*1/2*1/2 + 1/4*1/2 + 1/8
= 3/8

Why is it giving me a wrong answer?I know I am missing something.

Can someone please explain?

Infact Is my methos wrong, coz i am trying to solve Qs by this method. I understood the 1-p methos as well.


Direct probability approach:
One store does not have an iced tea: P(YYN)=3*1/2*1/2^2=3/8 (multiplying by 3 as scenario YYN can occur in 3!/2!=3 ways: YYN, YNY, NYY);
Two shops does not have an iced tea: P(YNN)=3*1/2^2*1/2=3/8 (multiplying by 3 as scenario YNN can occur in 3!/2!=3 ways: YNN, NYN, NNY);
All three shops does not have an iced tea: P(NNN)=1/2^3=1/8;

P=3/8+3/8+1/8=7/8.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Intern
Intern
Joined: 12 Oct 2012
Posts: 17
WE: General Management (Hospitality and Tourism)
Followers: 0

Kudos [?]: 8 [0], given: 38

Re: probability [#permalink] New post 17 Nov 2012, 07:05
Bunuel wrote:
aditi2013 wrote:
Bunuel wrote:
1. 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 be able to buy a can of iced tea?

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 P=1-(none of the shops has an iced tea)=1-P(NNN)=1-1/2^3=7/8.

2. The probability that a convenience store has no iced tea is 50%. If James stops by 3 convenience stores on his way to work, what is the probability that at least one of the stores will not have a can of iced tea?

The probability that at least one of the stores will not have a can of iced tea is 1-(all shops have an iced tea)=1-P(YYY)=1-1/2^2=7/8. Notice that we have the same answer as above. That's because the probability distribution is symmetrical for this question thus the probability that at least one of the stores will NOT have a can of iced tea = the probability that at least one of the stores will have a can of iced tea

3. 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?

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.

Hope it's clear.



Hi Bunuel,

WRT Q2.


P(atleast one does not have)= P(one does not have) + P( two don't have) + P(all 3 don't have)

= P(one does not have)*P(2 have) +
P (2 don't have)*P(one has) +
P(all 3 don't have)
= 1/2*1/2*1/2 + 1/4*1/2 + 1/8
= 3/8

Why is it giving me a wrong answer?I know I am missing something.

Can someone please explain?

Infact Is my methos wrong, coz i am trying to solve Qs by this method. I understood the 1-p methos as well.


Direct probability approach:
One store does not have an iced tea: P(YYN)=3*1/2*1/2^2=3/8 (multiplying by 3 as scenario YYN can occur in 3!/2!=3 ways: YYN, YNY, NYY);
Two shops does not have an iced tea: P(YNN)=3*1/2^2*1/2=3/8 (multiplying by 3 as scenario YNN can occur in 3!/2!=3 ways: YNN, NYN, NNY);
All three shops does not have an iced tea: P(NNN)=1/2^3=1/8;

P=3/8+3/8+1/8=7/8.

Hope it's clear.


Thanks Bunuel. I am really glad that I am a GC member.

Just one last Question, how do we come to know that we have to take the scenarios of possible positions of the outcomes (like yyn, yny, nyy)??
I often miss on that. Also is there a general rule which applies here to take the positions into account.
I have read all theory but guess challenges come only when u start solving. Pls help.
Re: probability   [#permalink] 17 Nov 2012, 07:05
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