Find the smallest number of people you need to choose
at random so that the probability that at least two of them
were both born on April $1$ exceeds $1/2$.
My answer:
Total of n people chosen at random
Event E -> At least $2$ people among the n people have birthday on April $1$
Event E' -> Only $1$ person has birthday on April 1 and $n-1$ people do NOT have birth day on April $1$
The probability for a person to have birthday on April 1 = $\frac{1}{365}$
The probability for a person to NOT have birthday on April 1 = $\frac{364}{365}$
$P(E)=1-P(E') = 1-((\frac{364}{365})^{n-1}) > \frac{1}{2}$
Solving the inequality gives the smallest $n = 254$
However, Book answer is $614$.
How ? Can someone please explain?
Best Answer
Let us add the event $E''$ of no person having a birthday on April 1, like suggested by 5xum. Then $$ \begin{align} P(E')&=\frac{1}{365}\cdot\left(\frac{364}{365}\right)^{n-1}\cdot\binom{n}{1}\\ &=\frac{364^{n-1}}{365^n}\cdot n\\ &\text{ }\\ &\text{ }\\ P(E'')&=\left(\frac{364}{365}\right)^{n}\cdot\binom{n}{0}\\ &=\frac{364^n}{365^n} \end{align} $$ And when you then solve $P(E)=1-P(E')-P(E'')=1/2$ you get $n\approx 612.257$ so for $n\geq 613$ you get the desired inequality. Link to solution using Wolfram Alpha.