GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Oct 2019, 23:25

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

Probability that the birthdays of 6 different persons will fall in exa

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
User avatar
Joined: 03 Mar 2010
Posts: 347
Schools: Simon '16 (M$)
Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 28 Apr 2011, 05:42
18
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

23% (02:05) correct 77% (02:01) wrong based on 68 sessions

HideShow timer Statistics

Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is

A. 341/12^6
B. 352/12^6
C. 352/12^5
D. 341/12^5
E. 371/12^6

_________________
My dad once said to me: Son, nothing succeeds like success.
Most Helpful Community Reply
Retired Moderator
avatar
Joined: 20 Dec 2010
Posts: 1571
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 28 Apr 2011, 10:05
5
1
subhashghosh wrote:
@fluke, please explain this :

000000,111111(All Birthdays in the same month)


6 people have two favorable choices each; Let's say the choices were 0(JAN),1(FEB) for simplicity just as it would be for head and tail.

Thus, total number of combinations would be: 2^6.

000000
000001
000010
000011
.
.
111111

Total=2^6

As per the condition that all 6 must have EXACTLY two calender months as their birthdays, at least two people should have different birthdays. They all can't have 000000 i.e. JAN,JAN,JAN,JAN,JAN,JAN or 1,1,1,1,1,1 i.e. FEB,FEB,FEB,FEB,FEB,FEB.

So, Favorable possibilities = 2^6-2=62.
_________________
General Discussion
Retired Moderator
avatar
Joined: 20 Dec 2010
Posts: 1571
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 28 Apr 2011, 07:32
4
jamifahad wrote:
Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is

A. 341/12^6
B. 352/12^6
C. 352/12^5
D. 341/12^5
E. 371/12^6


Sol:

Assuming that any person having a birthday on any month to be \(\frac{1}{12}\). In reality; probability of a person having a birthday on Feb(28 days) is less than having a birthday on January(31 days).

\(C^{12}_2*\frac{2^6-2}{12^6}=\frac{341}{12^5}\)

\(C^{12}_2\) - Number of ways to choose 2 months out of 12
\(2^6-2\) - 2 favorable choices(months) available for every person excluding 000000,111111(All Birthdays in the same month)
\(12^6\) - Every person has 12 Total possible months

Ans: "D"
_________________
Retired Moderator
avatar
B
Joined: 16 Nov 2010
Posts: 1245
Location: United States (IN)
Concentration: Strategy, Technology
Reviews Badge
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 28 Apr 2011, 09:54
@fluke, please explain this :

000000,111111(All Birthdays in the same month)
_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 28 Apr 2011, 18:13
1
jamifahad wrote:
Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is

A. 341/12^6
B. 352/12^6
C. 352/12^5
D. 341/12^5
E. 371/12^6


Another slightly different way of approaching this (and a little more complicated than fluke's - which I think is a very appropriate solution. Nevertheless, it might be useful for people who have done permutation combination before).
Select 2 months and then no of people for each month. You select two months in 12C2 ways. Say you get J and F.
Then you say that J could have 1 person's bday, 2 people's bday, 3 people's bday etc

J - 1/2/3/4/5 (You select 1 or 2 or 3 or 4 or 5 people for J)
F - 5/4/3/2/1 (Corresponding to no of people selected for J, remaining people go in F)

The distribution can be done in \(6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 2^6 - 2 = 62\) ways (If you recognize it, great, else just calculate)

So probability \(= 12C2 * 62/ 12^6 = 341/12^5\)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern
Intern
User avatar
Status: It's a long road ahead
Joined: 05 Feb 2011
Posts: 3
Location: Chennai, India
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 28 Apr 2011, 23:43
VeritasPrepKarishma wrote:
jamifahad wrote:
Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is

A. 341/12^6
B. 352/12^6
C. 352/12^5
D. 341/12^5
E. 371/12^6


Another slightly different way of approaching this (and a little more complicated than fluke's - which I think is a very appropriate solution. Nevertheless, it might be useful for people who have done permutation combination before).
Select 2 months and then no of people for each month. You select two months in 12C2 ways. Say you get J and F.
Then you say that J could have 1 person's bday, 2 people's bday, 3 people's bday etc

J - 1/2/3/4/5 (You select 1 or 2 or 3 or 4 or 5 people for J)
F - 5/4/3/2/1 (Corresponding to no of people selected for J, remaining people go in F)

The distribution can be done in \(6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 2^6 - 2 = 62\) ways (If you recognize it, great, else just calculate)

So probability \(= 12C2 * 62/ 12^6 = 341/12^5\)



My understanding of your solution goes this way.Let me know if I'm wrong.
1. Selecting any two months out of the 12 available can be done in 12C2 ways.
2. The 6 days can be distributed in the 2 available months in the following ways

M1 M2
0 6
1 5
2 4
3 3
4 2
5 1
6 0

which essentially is 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6,
which is of the form
(x+a)^6 = x^6 + x^5 6C1 a^1 + x^4 6C2 a^2 + x^3 6C3 a^3 + . . + a^6.
Substituting 1 for 'x' and 'a' we arrive at (1+1)^6 = 1 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 1
2^6 = 2 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5
i.e, 2^6-2 = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 62
Here, we subtract 2 because we are not to take into account either cases where all 6 birthdays fall in M1 or M2

Thus the probability becomes 12C2 * 62/ 12^6 = 341/12^5
_________________
Deepakh
Director
Director
avatar
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 653
Reviews Badge
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 28 Apr 2011, 23:59
Exactly as the binomial dist you have written! That is basically weeding out all the cases which have "all" 6 birthdays in one month i.e. 6C6 and 6C0
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 29 Apr 2011, 04:21
1
deepakh88 wrote:

My understanding of your solution goes this way.Let me know if I'm wrong.
1. Selecting any two months out of the 12 available can be done in 12C2 ways.
2. The 6 days can be distributed in the 2 available months in the following ways

M1 M2
0 6
1 5
2 4
3 3
4 2
5 1
6 0

which essentially is 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6,
which is of the form
(x+a)^6 = x^6 + x^5 6C1 a^1 + x^4 6C2 a^2 + x^3 6C3 a^3 + . . + a^6.
Substituting 1 for 'x' and 'a' we arrive at (1+1)^6 = 1 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 1
2^6 = 2 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5
i.e, 2^6-2 = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 62
Here, we subtract 2 because we are not to take into account either cases where all 6 birthdays fall in M1 or M2

Thus the probability becomes 12C2 * 62/ 12^6 = 341/12^5


Yes, that's correct. Many of us have used binomial theorem extensively and know that:
\(nC0 + nC1 + nC2 + ..... nCn = 2^n\) (an application of binomial as described by you)
Hence, it can be done this way.

On the other hand, you could obviously say that each of the six people have 2 options - either his bday falls in J or in F. So those are a total of \(2*2*2*2*2*2 = 2^6\) ways. Out of those, we need to remove 2 ways: All bdays fall in J and all bdays fall in F. Hence it can be done in \((2^6 - 2)\) ways.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Director
Director
User avatar
D
Affiliations: IIT Dhanbad
Joined: 13 Mar 2017
Posts: 728
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 05 Sep 2017, 00:15
jamifahad wrote:
Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is

A. 341/12^6
B. 352/12^6
C. 352/12^5
D. 341/12^5
E. 371/12^6



Really mind boggling question, You can't solve it in one go unless and until you have solved such question before..
fluke & VeritasPrepKarishma very nice solution...
It seems easy once we read the solution but it's a real tough question..
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu


Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)



What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 10 Sep 2017, 01:49
1
VeritasPrepKarishma wrote:
jamifahad wrote:
Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is

A. 341/12^6
B. 352/12^6
C. 352/12^5
D. 341/12^5
E. 371/12^6


Another slightly different way of approaching this (and a little more complicated than fluke's - which I think is a very appropriate solution. Nevertheless, it might be useful for people who have done permutation combination before).
Select 2 months and then no of people for each month. You select two months in 12C2 ways. Say you get J and F.
Then you say that J could have 1 person's bday, 2 people's bday, 3 people's bday etc

J - 1/2/3/4/5 (You select 1 or 2 or 3 or 4 or 5 people for J)
F - 5/4/3/2/1 (Corresponding to no of people selected for J, remaining people go in F)

The distribution can be done in \(6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 2^6 - 2 = 62\) ways (If you recognize it, great, else just calculate)

So probability \(= 12C2 * 62/ 12^6 = 341/12^5\)


Responding to a pm:
Quote:
How the calculation came to 341 as (12C2 ) 66*31 is not 341. Please help.

12C2 = 12*11/2

\(\frac{12C2 * 62}{12^6} = \frac{12*11*62}{2*12^6} = \frac{11*31}{12^5} = \frac{341}{12^5}\)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern
Intern
avatar
Joined: 02 Jan 2018
Posts: 1
Re: Probability that the birthdays of 6 different persons will fall in exa  [#permalink]

Show Tags

New post 02 Jan 2018, 18:21
How did n(s) become 12*6

Posted from my mobile device
GMAT Club Bot
Re: Probability that the birthdays of 6 different persons will fall in exa   [#permalink] 02 Jan 2018, 18:21
Display posts from previous: Sort by

Probability that the birthdays of 6 different persons will fall in exa

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne