Find the smallest number of people you need to choose
at random so that the probability that at least one of them
has a birthday today exceeds 1/2.
My Approach/Attempt-:
$P\left ( No\, \, one\, \, born \, \, today \right )+P\left ( atleast \, \, one\, \, born\, \, today \right )=1$
$P\left ( atleast\, \, one\, \, born \, \, today \right )=1-P\left ( No\, \, one\, \, born\, \, today \right )$
$P\left ( atleast \, \, one \, \, born\, \, today \right )$=$1-\left ( 364/365 \right )^{n}$
Answer will be $1-\left ( 364/365 \right )^{n}\geq 1/2$
how to solve $1-\left ( 364/365 \right )^{n}\geq 1/2$ and then find n?
Best Answer
A few steps: $$1−(364/365)^n\geq 1/2$$ $$1/2=1−1/2\geq (364/365)^n$$ $$\log(1/2)\geq n\log(364/365)$$ $$\log(2)\leq n(\log(365/364))$$ Can you take it form here?