BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

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.
Join the discussion
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
Image
Join the discussion

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
Join the discussion

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
Image
Join the discussion