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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 = 1/365

The probability for a person to NOT have birthday on April 1 = 364/365

$P(E)=1-P(E') = 1-((364/365)^{n-1}) > 1/2$

Solving the inequality gives the smallest n = 254

However, Book answer is 614.

How ?? Can someone please explain?

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    $\begingroup$ You forgot about the event "no person has a birthday on april 1." $\endgroup$ – 5xum Jul 29 '14 at 8:02
  • $\begingroup$ You forgot loads of details! Binomial coefficients?! I get the result 613, however ... $\endgroup$ – String Jul 29 '14 at 9:07
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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.

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  • $\begingroup$ Doing this for a leap year gives the book answer, 614 $\endgroup$ – Sakthi Jul 29 '14 at 14:16

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