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.
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.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
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
