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# How many randomly assembled people are needed to have a

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Intern
Joined: 26 Sep 2017
Posts: 23
GMAT 1: 640 Q48 V30
Re: How many randomly assembled people are needed to have a  [#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: 52344
Re: How many randomly assembled people are needed to have a  [#permalink]

### Show Tags

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: 26 Sep 2017
Posts: 23
GMAT 1: 640 Q48 V30
Re: How many randomly assembled people are needed to have a  [#permalink]

### Show Tags

10 Apr 2018, 23:11
Bunuel wrote:
Bakshi121092 wrote:
Bunuel wrote:
[quote="Bakshi121092"]

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).[/quote]Sure I'll look into it. I get the understanding now.
Thanks for the help.

Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app
Re: How many randomly assembled people are needed to have a &nbs [#permalink] 10 Apr 2018, 23:11

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