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# How many randomly assembled people are needed to have a

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Senior Manager
Joined: 25 Jun 2009
Posts: 289
How many randomly assembled people are needed to have a [#permalink]

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17 Jan 2010, 06:50
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53% (00:57) correct 47% (01:06) wrong based on 234 sessions

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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
[Reveal] Spoiler: OA
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Joined: 02 Sep 2009
Posts: 44627

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17 Jan 2010, 12:18
5
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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.

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10 Jan 2012, 06:41
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)

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Manager
Joined: 26 Apr 2013
Posts: 50
Location: United States
Concentration: Marketing, Nonprofit
GPA: 3.5
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Re: How many randomly assembled people are needed to have a [#permalink]

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03 Oct 2013, 18:57
cipher 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

Hi Bunuel,

Don't you think that this question is faulty?

The question ask about the probability better than 50%. If we have 4 people we still have probability better than 50% and if we chose 5 people it will be even more. So how can be the answer C
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Re: How many randomly assembled people are needed to have a [#permalink]

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01 Oct 2014, 12:00
1
KUDOS
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
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Re: How many randomly assembled people are needed to have a [#permalink]

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02 Oct 2014, 00:59
2
KUDOS
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
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Re: How many randomly assembled people are needed to have a [#permalink]

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12 May 2015, 01:54
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
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Re: How many randomly assembled people are needed to have a [#permalink]

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12 May 2015, 03:48
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.
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Re: How many randomly assembled people are needed to have a [#permalink]

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20 May 2015, 02:50
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.
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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]

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05 Aug 2015, 22:54
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Expert's post
1
This post was
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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%.

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews GMAT Tutor Joined: 24 Jun 2008 Posts: 1345 Re: What is the minimum number of randomly chosen people needed in order t [#permalink] ### Show Tags 07 Aug 2015, 21:42 1 This post received KUDOS Expert's post 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 SVP Joined: 08 Jul 2010 Posts: 2067 Location: India GMAT: INSIGHT WE: Education (Education) Re: What is the minimum number of randomly chosen people needed in order t [#permalink] ### Show Tags 08 Aug 2015, 00:27 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 Answer: option C _________________ 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 22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION Manager Joined: 04 May 2015 Posts: 72 Concentration: Strategy, Operations WE: Operations (Military & Defense) Re: What is the minimum number of randomly chosen people needed in order t [#permalink] ### Show Tags 09 Aug 2015, 10:21 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%. Answer (C) #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. Thanks in advance! _________________ If you found my post useful, please consider throwing me a Kudos... Every bit helps Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8030 Location: Pune, India Re: What is the minimum number of randomly chosen people needed in order t [#permalink] ### Show Tags 09 Aug 2015, 21:42 Expert's post 1 This post was BOOKMARKED 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%. Answer (C) #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. Thanks in advance! 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: What is the minimum number of randomly chosen people needed in order t [#permalink]

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09 Aug 2015, 22:01
1
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Expert's post
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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Retired Moderator Joined: 18 Sep 2014 Posts: 1183 Location: India Re: What is the minimum number of randomly chosen people needed in order t [#permalink] ### Show Tags 10 Aug 2015, 03:54 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 only time you can lose is when you give up. Try hard and you will suceed. Thanks = Kudos. Kudos are appreciated http://gmatclub.com/forum/rules-for-posting-in-verbal-gmat-forum-134642.html When you post a question Pls. Provide its source & TAG your questions Avoid posting from unreliable sources. My posts http://gmatclub.com/forum/beauty-of-coordinate-geometry-213760.html#p1649924 http://gmatclub.com/forum/calling-all-march-april-gmat-takers-who-want-to-cross-213154.html http://gmatclub.com/forum/possessive-pronouns-200496.html http://gmatclub.com/forum/double-negatives-206717.html http://gmatclub.com/forum/the-greatest-integer-function-223595.html#p1721773 https://gmatclub.com/forum/improve-reading-habit-233410.html#p1802265 SVP Joined: 08 Jul 2010 Posts: 2067 Location: India GMAT: INSIGHT WE: Education (Education) Re: What is the minimum number of randomly chosen people needed in order t [#permalink] ### Show Tags 10 Aug 2015, 04:31 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 22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8030 Location: Pune, India Re: What is the minimum number of randomly chosen people needed in order t [#permalink] ### Show Tags 10 Aug 2015, 22:08 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: How many randomly assembled people are needed to have a [#permalink]

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10 Apr 2018, 22:06
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.
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Posts: 44627
Re: How many randomly assembled people are needed to have a [#permalink]

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10 Apr 2018, 22:13
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 ! ! !
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Re: How many randomly assembled people are needed to have a   [#permalink] 10 Apr 2018, 22:13

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