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?

  • 5
    $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.