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

 It is currently 20 Jun 2019, 12:31 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

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.  How many randomly assembled people are needed to have a better than 50

Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 55732
How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

1
6 00:00

Difficulty:   65% (hard)

Question Stats: 49% (01:31) correct 51% (01:22) wrong based on 347 sessions

HideShow timer Statistics

How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 55732
How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

12
1
17
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.

Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.

So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$

For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$

Thus min 3 people are needed.

_________________
General Discussion
Manager  Joined: 11 Sep 2009
Posts: 127
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

1
1

Essentially, you need to find the number of people so that the probability of all of them not being from a leap year is LESS THAN 50%. The probability of any one person not being born on a leap year is approximately 3/4.

So you are essentially solving for:

$$(\frac{3}{4})^n < \frac{1}{2}$$

n = 1: 3/4 < 1/2? NO
n = 2: (3/4)(3/4) = 9 / 16 < 1/2? NO
n = 3 (3/4)(3/4)(3/4) = 27 / 64 < 1/2? YES

Senior Manager  Joined: 31 Oct 2011
Posts: 478
Schools: Johnson '16 (M)
GMAT 1: 690 Q45 V40 WE: Asset Management (Mutual Funds and Brokerage)
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

1
1
My solution

Prob(at least one of them was born in a leap year) > 50%
P(A) = 1 - P(A')
1 - Prob(none were born in a leap year) > 1/2
Prob(none) - 1 < -1/2
Prob(none) < 1/2

Assume that there are n people.
Prob(none) = (3/4)^n
(3/4)^n < 1/2
3^n < (4^n)/2
3^n < (2^2n)/2
3^n < 2^(2n-1)

If n = 3, 3^3 < 2^5 is true (27 < 32)

_________________
My Applicant Blog: http://hamm0.wordpress.com/
Intern  Joined: 07 Aug 2013
Posts: 5
Schools: Alberta FT '16
GMAT 1: 640 Q48 V31 Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

1
This is how I would ans .

P(people born in leap year) = 1/4 (because out of every 4 years 1 is a leap year )
P(People not born in leap year ) = 3/4

Now for any n no. of people (3/4)^n is Probability of those people who were not born in leap year

As total probability is 1

1-(3/4)^n represents the cumulative probability of all those people born in leap year ( Eg this P includes probability of 1 person in group to be
born in leap year to n persons in a group to be born in leap year - which ans at least 1 person is born in leap year )

So 1-(3/4)^n>1/2

after trying diff values of n , it comes out as 3
Manager  Joined: 21 Sep 2012
Posts: 213
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

3
Probability of a person to be born in a leap year = 1/4 ... <1/2
Probability of atleast one person out of two to be born in a leap year = 1/4+1/4=1/2 .... not >1/2
Probability of atleast one person out of 3 to be born in a leap year = 1/4+1/4+1/4=3/4..... >1/2

Hence minimum 3 people are required.
Ans = C
Manager  B
Joined: 14 Mar 2014
Posts: 145
GMAT 1: 710 Q50 V34 Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

Bunuel wrote:
nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

I found this question interesting, so thought of sharing it you folks.

Cheers

Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.

Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.

So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$

For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$

Thus min 3 people are needed.

Bunuel
how is Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$. ???

if you consider years 1896 1897 1898 1899 1900 1901 1902 1903 1904 .... Here 1896 and 1904 are the only leap years.
How did you generalize probability not being a leap year to be 3/4 ?
And also question didn't ask about minimum number of people needed. Ans can be 3 or 4 or 5
_________________
I'm happy, if I make math for you slightly clearer
And yes, I like kudos
¯\_(ツ)_/¯ Math Expert V
Joined: 02 Sep 2009
Posts: 55732
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

VenoMftw wrote:
Bunuel wrote:
nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

I found this question interesting, so thought of sharing it you folks.

Cheers

Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.

Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.

So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$

For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$

Thus min 3 people are needed.

Bunuel
how is Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$. ???

if you consider years 1896 1897 1898 1899 1900 1901 1902 1903 1904 .... Here 1896 and 1904 are the only leap years.
How did you generalize probability not being a leap year to be 3/4 ?
And also question didn't ask about minimum number of people needed. Ans can be 3 or 4 or 5

Yes, every 100's year is not leap but for this question it's implied that 1 in 4 years is leap.
_________________
Intern  Joined: 17 May 2015
Posts: 3
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

2
2
If you have 1 person, the chance that he's born in a leap year is $$\frac{1}{4}$$
If you have 2 persons the chances are $$\frac{2}{4}$$ = 50%
We need AT LEAST 50%, so 1 extra person will do the job.

We need 3 persons.

C.
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9340
Location: Pune, India
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

2
2
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Keeping it simple:

P (A person was born in leap year) = 1/4
This is less than 50% - not the answer

P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)

The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)

P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16

This is less than 50% but close - not the answer

So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.

If you want to calculate it, you can do it as given below:

P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)

This is definitely more than 50%.

_________________
Karishma
Veritas Prep GMAT Instructor

GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1634
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

1
1
Or we can reverse the problem. If the probability someone is born in a leap year is greater than 1/2, then the probability no one was born in a leap year needs to be less than 1/2.

If we pick one person, the probability they weren't born in a leap year is 3/4. If we pick two people, the probability both were not born in a leap year is (3/4)(3/4) = (3/4)^2 = 9/16. That's still bigger than 1/2, but if we pick a third person, the probability none of the three people was born in a leap year is (3/4)^3 which is less than 1/2, which means the probability someone was born in a leap year is now greater than 1/2. So the answer is three.
_________________
GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2943
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

1
1
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Every 4 consecutive years have 1 Leap year and 3 normal years

i.e. Probability of an year to be leap year = 1/4
and Probability of an year to NOT to be leap year = 1/4

Probability of 1 randomly selected year to be NON leap year = (3/4) i.e. Greater than 50%

Probability of 2 randomly selected year to be NON leap year = (3/4)*(3/4) = 9/16 i.e. Greater than 50%

Probability of 3 randomly selected year to be NON leap year = (3/4)*(3/4)*(3/4) = 27/64 i.e. Less than 50% i.e. there are more than 50% chances that atleast one of the selected 3 years will be a Leap Year

Hence - 3 Years

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Manager  Joined: 04 May 2015
Posts: 71
Concentration: Strategy, Operations
WE: Operations (Military & Defense)
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

VeritasPrepKarishma wrote:
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Keeping it simple:

P (A person was born in leap year) = 1/4
This is less than 50% - not the answer

I understand and follow this part

P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)

Because we are looking for "at least one" shouldn't we be trying to determine P(A, or B, or Both) born in a leap year? I had the same answer as you and had the correct answer but probabilities are a bit of a weakness of mine so trying to work on it. I will show below my working.

The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)

P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16

This is less than 50% but close - not the answer

So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.

If you want to calculate it, you can do it as given below:

P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)

This is definitely more than 50%.

#If one is selected P(A born in a leap year) = 1/4 or 25%

#If two are selected the probability that at least one is born in a leap year P(A only, B only, or A & B) = P(A) + P(B) + P(A&B) = (1/4 * 3/4) + (3/4 * 1/4) + (1/4 * 1/4) = 7/16

From here I just made the assumption that 3 people would push it above 50% and selected C.

I am just curious of the method that VeritasPrepKarishma used... Maybe there is a Thread or Blog post you could point me to that explains this in more detail.

_________________
If you found my post useful, please consider throwing me a Kudos... Every bit helps Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9340
Location: Pune, India
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

DropBear wrote:
VeritasPrepKarishma wrote:
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Keeping it simple:

P (A person was born in leap year) = 1/4
This is less than 50% - not the answer

I understand and follow this part

P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)

Because we are looking for "at least one" shouldn't we be trying to determine P(A, or B, or Both) born in a leap year? I had the same answer as you and had the correct answer but probabilities are a bit of a weakness of mine so trying to work on it. I will show below my working.

The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)

P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16

This is less than 50% but close - not the answer

So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.

If you want to calculate it, you can do it as given below:

P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)

This is definitely more than 50%.

#If one is selected P(A born in a leap year) = 1/4 or 25%

#If two are selected the probability that at least one is born in a leap year P(A only, B only, or A & B) = P(A) + P(B) + P(A&B) = (1/4 * 3/4) + (3/4 * 1/4) + (1/4 * 1/4) = 7/16

From here I just made the assumption that 3 people would push it above 50% and selected C.

I am just curious of the method that VeritasPrepKarishma used... Maybe there is a Thread or Blog post you could point me to that explains this in more detail.

This method is a counterpart of SETS formulas we use.

10 people like A and 20 people like B. 5 like both A and B. So how many people like A or B?

n(A or B) = n(A) + n(B) - n(A and B)
We subtract n(A and B) because it is counted twice - once in n(A) and another time in n(B). But we want to count it only once so subtract it out once.
n(a or B) gives us the number of people who like at least one of A and B.

Similarly, we can use this method for probability
p(A or B) = p(A) + p(B) - p(A and B)

It is exactly the same concept. It gives us the probability that at least one of two people are born in a leap year.
When considering p(A), p(B) is to be ignored and when c considering p(B), p(A) is to be ignored. We take care of both when we subtract p(A and B).

Similarly, we can use the sets concept for 3 people using the three overlapping sets formula:

n(A or B or C) = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)

it becomes
p(A or B or C) = p(A) + p(B) + p(C) - p(A and B) - p(B and C) - p(C and A) + p(A and B and C)
_________________
Karishma
Veritas Prep GMAT Instructor

Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9340
Location: Pune, India
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

1
DropBear wrote:
VeritasPrepKarishma wrote:
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Keeping it simple:

P (A person was born in leap year) = 1/4
This is less than 50% - not the answer

I understand and follow this part

P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)

Because we are looking for "at least one" shouldn't we be trying to determine P(A, or B, or Both) born in a leap year? I had the same answer as you and had the correct answer but probabilities are a bit of a weakness of mine so trying to work on it. I will show below my working.

The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)

P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16

This is less than 50% but close - not the answer

So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.

If you want to calculate it, you can do it as given below:

P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)

This is definitely more than 50%.

#If one is selected P(A born in a leap year) = 1/4 or 25%

#If two are selected the probability that at least one is born in a leap year P(A only, B only, or A & B) = P(A) + P(B) + P(A&B) = (1/4 * 3/4) + (3/4 * 1/4) + (1/4 * 1/4) = 7/16

From here I just made the assumption that 3 people would push it above 50% and selected C.

I am just curious of the method that VeritasPrepKarishma used... Maybe there is a Thread or Blog post you could point me to that explains this in more detail.

Also, check out this blog post: http://www.veritasprep.com/blog/2012/01 ... e-couples/

Here I have used sets in a combination problem. Conceptually they are all the same.
_________________
Karishma
Veritas Prep GMAT Instructor

Retired Moderator S
Joined: 18 Sep 2014
Posts: 1102
Location: India
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

good question. But i doubt whether we can solve like this

#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two is selected P(A,B born in a leap year) = 2/4

Since we need more than 50% we pickup 3 as it is minimal among others.
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2943
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

Mechmeera wrote:
good question. But i doubt whether we can solve like this

#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two is selected P(A,B born in a leap year) = 2/4

Since we need more than 50% we pickup 3 as it is minimal among others.

Hi Mechmeera,

Highlighted step is the wrong step that you have written

Because when you select two individuals then one of them can be born in Leap Year and other is not and vise versa as well

So Probability of both A and B born in Leap year = (1/4)for A * (1/4)for B = 1/16

But Probability of One of them Born in leap year = (1/4)*(3/4) + (1/4)*(3/4) = 6/16

i.e. Probability that atleast one of A and B born in Leap year = (1/16)+(6/16) = 7/16 which is LESS than 50%

But when you follow the same process for 3 people then the probability exceeds 50%

I hope it helps!
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9340
Location: Pune, India
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

Mechmeera wrote:
good question. But i doubt whether we can solve like this

#If one is selected P(A born in a leap year) = 1/4 or 25%
#If two is selected P(A,B born in a leap year) = 2/4

Since we need more than 50% we pickup 3 as it is minimal among others.

The thought process is correct but the execution is not.
The method I have used is the same as this. When two are selected, the probability that at least one of them has bday in a leap year is 2/4 - 1/16, not only 2/4. I have explained 'why' here: what-is-the-minimum-number-of-randomly-chosen-people-needed-in-order-t-203018.html#p1559070
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 26 Sep 2017
Posts: 22
GMAT 1: 640 Q48 V30 Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

Bunuel wrote:
nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

I found this question interesting, so thought of sharing it you folks.

Cheers

Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.

Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.

So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$

For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$

Thus min 3 people are needed.

Hi Bunuel,

What is wrong in the following approach?

Cases considered :
1. Born in a leap year
2. Not born in a leap year

In that case, the probability of not born in a leap year =1/2

And subsequently, the probability that at least one of them is born in a leap year would turn out to be greater than 50% for 2 people.

Thank you.
Math Expert V
Joined: 02 Sep 2009
Posts: 55732
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

Show Tags

Bakshi121092 wrote:
Bunuel wrote:
nitishmahajan wrote:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

I found this question interesting, so thought of sharing it you folks.

Cheers

Probability of a randomly selected person NOT to be born in a leap year is $$\frac{3}{4}$$.

Among 2 people, probability that none of them was born in a leap is $$\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$$. The probability at least one born in leap from these two is $$1- \frac{9}{16}=\frac{7}{16}<\frac{1}{2}$$.

So, we are looking for such n (# of people), when $$1-(\frac{3}{4})^n>\frac{1}{2}$$

For $$n=3$$ --> $$1-\frac{27}{64}=\frac{37}{64}>\frac{1}{2}$$

Thus min 3 people are needed.

Hi Bunuel,

What is wrong in the following approach?

Cases considered :
1. Born in a leap year
2. Not born in a leap year

In that case, the probability of not born in a leap year =1/2

And subsequently, the probability that at least one of them is born in a leap year would turn out to be greater than 50% for 2 people.

Thank you.

Let me ask you: is the probability that you win a lottery 1/2?
1. You win the lottery
2. You won't win the lottery

Or: what is the probability to meet a dinosaur on the street? Is it 1/2? Either you meet it or not?

22. Probability

For more:
ALL YOU NEED FOR QUANT ! ! !
_________________ Re: How many randomly assembled people are needed to have a better than 50   [#permalink] 10 Apr 2018, 22:13

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by

How many randomly assembled people are needed to have a better than 50  