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Probability

Expert replies
by rakeshd347 » Wed Oct 02, 2013 6:04 am
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

OA soon.
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Source: — Problem Solving |

by Brent@GMATPrepNow » Wed Oct 02, 2013 6:37 am
rakeshd347 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

OA soon.
Aside: I don't the GMAT would require someeone to know what a leap year is.

Let's say P(born in a leap year) = 1/4 (approximately).
So, P(not born in a leap year) = 3/4

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(at least 1 born in a leap year) = 1 - P(not at least 1 born in a leap year)
= 1 - P(zero born in a leap year)

Let's test a few values.

2 people
P(zero born in a leap year) = P(1st person is not leap year AND 2nd person is not leap year)
= P(1st person is not leap year) x P(2nd person is not leap year)
= (3/4) x (3/4)
= 9/16

So P(at least 1 born in a leap year) = 1 - P(zero born in a leap year)
= 1 - 9/16
= 7/16

7/16 is less than 1/2, so we see still need MORE people to get the probability over 1/2.
IMPORTANT: Since 7/16 is barely less than 1/2, we should see that one more person will do the trick, which means the correct answer is C

If we want to verify it, the calculations are as follows:

3 people
P(zero born in a leap year) = P(1st person is not leap year AND 2nd person is not leap year AND 3rd person is not leap year)
= P(1st person is not leap year) x P(2nd person is not leap year) x P(3rd person is not leap year)
= (3/4) x (3/4) x (3/4)
= 27/64

So P(at least 1 born in a leap year) = 1 - P(zero born in a leap year)
= 1 - 27/64
= 37/64

Since 37/64 is greater than 1/2, the correct answer is C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by mevicks » Wed Oct 02, 2013 7:10 am
Brent@GMATPrepNow wrote:
rakeshd347 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

OA soon.
Let's say P(born in a leap year) = 1/4 (approximately).
So, P(not born in a leap year) = 3/4
Hi Brent,

By this logic even the answer choices D and E yield a probability of greater than 50%.
D --> 1 - 81/256 --> 175/256
E --> 1 - 243/1024 --> 781/1024

Since the question doesn't place a lower limit on the number of people (least number of people required) why should one choose option c over the other two?

Thanks & Regards,
Vivek
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by Brent@GMATPrepNow » Wed Oct 02, 2013 7:15 am
mevicks wrote: Hi Brent,

By this logic even the answer choices D and E yield a probability of greater than 50%.
D --> 1 - 81/256 --> 175/256
E --> 1 - 243/1024 --> 781/1024

Since the question doesn't place a lower limit on the number of people (least number of people required) why should one choose option c over the other two?

Thanks & Regards,
Vivek
Good point, Vivek.
The question is poorly worded, so I guess I just instinctively added more words to make it more GMAT-like. As it is originally worded, C, D and E are all correct.

The question should have something that says, "What is the minimum number of randomly-selected people in order to have a better than 50% probability that at least 1 of them was born in a leap year?" (in which case the correct answer is C)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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