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

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Intern
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]

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10 Apr 2018, 21:43
Bunuel wrote:
Bakshi121092 wrote:

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 ! ! !

I understand. But the question doesn't mention a time frame.
For example, considering the probability to be 3/4 is for any 4 year time frame. If we consider a 5 year time frame, the probabilities change.

The probabilities you asked depend on other factors. One can simply not tell. But with no other information given, the probability that I could meet a dinosaur would be 1/2. I may or may not encounter a dinosaur.

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 61403
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

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10 Apr 2018, 22:35
Bakshi121092 wrote:
Bunuel wrote:
Bakshi121092 wrote:

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 ! ! !

I understand. But the question doesn't mention a time frame.
For example, considering the probability to be 3/4 is for any 4 year time frame. If we consider a 5 year time frame, the probabilities change.

The probabilities you asked depend on other factors. One can simply not tell. But with no other information given, the probability that I could meet a dinosaur would be 1/2. I may or may not encounter a dinosaur.

Thanks.

That's not correct.

There is one leap year in four. So, the probability of random person being born in leap year is 1/4. As for the dinosaur problem, no other information is needed apart from common knowledge, the probability is not 1/2: meeting and not meeting are not equally likely, and simply the fact that there are only two options does not mean that the provability is 1/2.

I suggest you to follow the links given in my previous post and maybe read some simple book on probability (introduction of some kind).
_________________
Intern
Joined: 22 Oct 2019
Posts: 27
Location: India
Schools: HEC Montreal '21
GMAT 1: 660 Q50 V30
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

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21 Jan 2020, 02:54
Bunuel 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

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.

How did we deduce the probability for any person to be 3/4? I did not understand that.
Math Expert
Joined: 02 Sep 2009
Posts: 61403
Re: How many randomly assembled people are needed to have a better than 50  [#permalink]

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21 Jan 2020, 02:58
ashita5678 wrote:
Bunuel 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

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.

How did we deduce the probability for any person to be 3/4? I did not understand that.

There is 1 leap year in 4 years. So, the probability of a person to be born in a leap year is 1/4. So, the probability that a person is NOT born in a leap year is 3/4.
_________________
Re: How many randomly assembled people are needed to have a better than 50   [#permalink] 21 Jan 2020, 02:58

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