There are 12 signs of the zodiac. How many people must be present for there to...

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)?

Solutions

Expert Solution

TOPIC:Classical definition of probability.

orchestra answered 2 years ago

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