Question

How many randomly chosen guests should I invite to my party so that the probability of having a guest with the same birthday as mine is at least 2/3?

Answer 1

Answer:

At least 401 chosen guests

Step-by-step explanation:

Represent those with same birth day with Same

So:

$P(Same) > \frac{2}{3}$

Represent number of people with n

There are 365 days in a year and 364 days out of these days is not your birthday.

So, there the probability that n people do not share your birthday is $(\frac{364}{365})^n$

i.e.

$P(none) = (\frac{364}{365})^n$

Solving further, we have that:

$P(same) + P(none) = 1$

$P(same) + (\frac{364}{365})^n = 1$

$P(same) = 1 - (\frac{364}{365})^n$

We're calculating the probability that $P(Same) > \frac{2}{3}$.

So, we have:

$1 - (\frac{364}{365})^n > \frac{2}{3}$

Collect Like Terms

$1 - \frac{2}{3}> (\frac{364}{365})^n$

$\frac{3-2}{3}> (\frac{364}{365})^n$

$\frac{1}{3}> (\frac{364}{365})^n$

$3^{-1} > (\frac{364}{365})^n$

Take natural logarithm (ln) of both sides

$ln(3^{-1}) > ln((\frac{364}{365})^n)$

$-ln\ 3 > n\ ln\ (\frac{364}{365})$

Apply laws of logarithm

$-ln\ 3 > n\ (ln(364) - ln(365))$

Multiply through by -1

$ln\ 3 < n\ (ln(365) - ln(364))$

Solve for n

$\frac{ln\ 3}{(ln(365) - ln(364))} < n$

Reorder

$n > \frac{ln\ 3}{(ln(365) - ln(364))}$

$n > \frac{1.09861228867}{(5.89989735358 - 5.89715386764)}$

$n> \frac{1.09861228867}{0.00274348594}$

$n > 400.443928891$

$n > 400$ (approximated)

This implies that n = 401, 402, 403 .....

i.e

$n \geq 401$

So, at least 401 people has to be invited