Question

There are 12 signs of the zodiac. How many people must be present for there to be at least a 50% chance that two or more of them were born under the same sign?

1-P(no one has the same sign) >= 0.5

P(two or more people have the same sign) = 1-P(no one has the same sign)

= 1 - (12!/(12-n)!)/12^n

1-0.5=0.5 >= P(no one has the same sign) = 12!/(12^n * (12-n)!)

0.5 >= 12!/(12^n * (12-n)!)

Solve for n.

n = 5

My question is, why is the probability that no one has the same sign? Could you explain in detail how do you arrive to (12!/(12-n)!)/12^n)?

Answer 1

TOPIC:Classical definition of probability.