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Leap year

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Leap year

by gmatblood » Fri Nov 04, 2011 1:16 pm
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

IMO: C
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by user123321 » Fri Nov 04, 2011 3:43 pm
gmatblood 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

IMO: C
lets assume n people are present.
the probability of one person having bday in leap year is 1/4.
the probabliity of n people of which atleast one of them born in leap year = 1- (probability of none of them born in leap year)
=> p = 1 - (3/4)^n > 1/2 (from given)
=> p = (3/4)^n<1/2
so when n is 3 it will be just less than half. So should be C.

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by satishchandra » Fri Nov 04, 2011 11:07 pm
Prob.that a person is born in leap year = 1/4 = 25%

We need atleast 3 or more people to have probability more than 50%

As it is given at least 1 of them was born in a leap year
We need 2 minumim to have probabiity to be more than 50%

My ans:B

However, I am not too OK with this wording? "How many randomly assembled people"
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by Anurag@Gurome » Fri Nov 04, 2011 11:10 pm
gmatblood 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

IMO: C
Probability of a randomly selected person to be born in a leap year = 1/4
Probability of a randomly selected person NOT to be born in a leap year = 1 - 1/4 = 3/4
Among 2 people, probability that none of them was born in a leap year = 3/4 * 3/4 = 9/16.
The probability that at least one of them was born in a leap year = 1 - 9/16 = 7/16 < 1/2
Now, for such n people, when 1-(3/4)^n > 1/2
n = 3 implies 1 - 27/64 = 37/64 > 1/2
So, minimum of 3 people are needed.

The correct answer is C.
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by satishchandra » Fri Nov 04, 2011 11:12 pm
Can someone explain What does "How many randomly assembled people"indicate?
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by GmatMathPro » Sat Nov 05, 2011 6:09 am
satishchandra wrote:Can someone explain What does "How many randomly assembled people"indicate?
For the calculations above to be valid, we have to assume that the probabilities are independent. That is, we have to assume that whether or not person A was born in a leap year in no way affects the probability of person B having been born in a leap year. This statement is what allows us to reasonably assume this. It's basically saying that the people are chosen randomly from the general population. Otherwise, the probabilities could be dependent. An extreme example would be if the two people under consideration were twins. In that case it would be insane to say the probability of them both not being born in a leap is is (3/4)(3/4)=9/16. Because if one of them was not born in a leap year, we can be virtually certain that the other one also was not. You'd run into similar problems if we were looking at people from the same high school class. So, to head off all these problems, they include this statement to clarify that the probabilities are essentially independent. If I had written this problem I probably would not have phrased it exactly this way, but I think it gets the job done.
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by vaibhavgupta » Sat Nov 05, 2011 8:22 am
gmatblood 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

IMO: C
1/4 +1/4+1/4= 3/4 >1/2
Hence C!!
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by GmatMathPro » Sat Nov 05, 2011 9:20 am
vaibhavgupta wrote:
gmatblood 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

IMO: C
1/4 +1/4+1/4= 3/4 >1/2
Hence C!!
Sorry, but this only works coincidentally. By this line of reasoning, if you had 5 randomly chosen people, the probability of at least one of them having been born in a leap year would be 5/4 or 125%.
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by vaibhavgupta » Sat Nov 05, 2011 9:21 am
GmatMathPro wrote:
vaibhavgupta wrote:
gmatblood 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

IMO: C
1/4 +1/4+1/4= 3/4 >1/2
Hence C!!
Sorry, but this only works coincidentally. By this line of reasoning, if you had 5 randomly chosen people, the probability of at least one of them having been born in a leap year would be 5/4 or 125%.
Ok i did not know this. :) could u explain a tad bit more?
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by GmatMathPro » Sat Nov 05, 2011 10:03 am
vaibhavgupta wrote:
GmatMathPro wrote:
vaibhavgupta wrote:
gmatblood 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

IMO: C
1/4 +1/4+1/4= 3/4 >1/2
Hence C!!
Sorry, but this only works coincidentally. By this line of reasoning, if you had 5 randomly chosen people, the probability of at least one of them having been born in a leap year would be 5/4 or 125%.
Ok i did not know this. :) could u explain a tad bit more?
For simplicity, let's focus on just two people. If the question is "What is the probability of at least one person being born in a leap year?" and you say the probability of person A being born in a leap year or person B being born in a leap year is 1/4+1/4=1/2, you are double counting the probability that they are both born in a leap year, so 1/2 is an overestimate of the true probability. If you do it this way you have to subtract out the probability that they are both born in a leap year: 1/4+1/4-(1/4)(1/4)=7/16. It gets much more complicated when we add more people. that's why the preferred way to calculate probabilities like these is to subtract the probability of it NOT happening from 1: 1-(3/4)(3/4)=1-9/16=7/16.
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by vaibhavgupta » Sat Nov 05, 2011 10:06 am
GmatMathPro wrote:
vaibhavgupta wrote:
GmatMathPro wrote:
vaibhavgupta wrote:
gmatblood 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

IMO: C
1/4 +1/4+1/4= 3/4 >1/2
Hence C!!
Sorry, but this only works coincidentally. By this line of reasoning, if you had 5 randomly chosen people, the probability of at least one of them having been born in a leap year would be 5/4 or 125%.
Ok i did not know this. :) could u explain a tad bit more?
For simplicity, let's focus on just two people. If the question is "What is the probability of at least one person being born in a leap year?" and you say the probability of person A being born in a leap year or person B being born in a leap year is 1/4+1/4=1/2, you are double counting the probability that they are both born in a leap year, so 1/2 is an overestimate of the true probability. If you do it this way you have to subtract out the probability that they are both born in a leap year: 1/4+1/4-(1/4)(1/4)=7/16. It gets much more complicated when we add more people. that's why the preferred way to calculate probabilities like these is to subtract the probability of it NOT happening from 1: 1-(3/4)(3/4)=1-9/16=7/16.
So, in such a scenario, we should always calculate of that even not happening?

I tend to do that when the calculation of probability of that event is highly to tough to calculate.
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by GmatMathPro » Sat Nov 05, 2011 10:17 am
vaibhavgupta wrote:
So, in such a scenario, we should always calculate of that even not happening?

I tend to do that when the calculation of probability of that event is highly to tough to calculate.
I would. And again, obviously you CAN calculate it directly, but with even three people it gets much more complicated, which means you have more opportunities to make a mistake. Of course, you'll want to make sure the probability you want is actually the same as 1-P(NOT).
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by vaibhavgupta » Sat Nov 05, 2011 10:31 am
GmatMathPro wrote:
vaibhavgupta wrote:
So, in such a scenario, we should always calculate of that even not happening?

I tend to do that when the calculation of probability of that event is highly to tough to calculate.
I would. And again, obviously you CAN calculate it directly, but with even three people it gets much more complicated, which means you have more opportunities to make a mistake. Of course, you'll want to make sure the probability you want is actually the same as 1-P(NOT).
Okay, would use this one then ! Thankss! :)
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by satishchandra » Sat Nov 05, 2011 9:22 pm
satishchandra wrote:Can someone explain What does "How many randomly assembled people"indicate?
GmatMathPro wrote: So, to head off all these problems, they include this statement to clarify that the probabilities are essentially independent. If I had written this problem I probably would not have phrased it exactly this way, but I think it gets the job done.
Thanks a lot MathPro. Formation of question is not ideal. Why to say randomly assembled? It should be randomly chosen or selected
Possibly the best method can be as under.

Equation = 1-(3/4)^n
if n=1; 25% < 50%
if n=2; 43% < 50%
if n=3; 57% > 50%

Hence, we need atleast 3 or more.
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