Answer:
At least 401 chosen guests
Step-by-step explanation:
Represent those with same birth day with Same
So:
[tex]P(Same) > \frac{2}{3}[/tex]
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 [tex](\frac{364}{365})^n[/tex]
i.e.
[tex]P(none) = (\frac{364}{365})^n[/tex]
Solving further, we have that:
[tex]P(same) + P(none) = 1[/tex]
[tex]P(same) + (\frac{364}{365})^n = 1[/tex]
[tex]P(same) = 1 - (\frac{364}{365})^n[/tex]
We're calculating the probability that [tex]P(Same) > \frac{2}{3}[/tex].
So, we have:
[tex]1 - (\frac{364}{365})^n > \frac{2}{3}[/tex]
Collect Like Terms
[tex]1 - \frac{2}{3}> (\frac{364}{365})^n[/tex]
[tex]\frac{3-2}{3}> (\frac{364}{365})^n[/tex]
[tex]\frac{1}{3}> (\frac{364}{365})^n[/tex]
[tex]3^{-1} > (\frac{364}{365})^n[/tex]
Take natural logarithm (ln) of both sides
[tex]ln(3^{-1}) > ln((\frac{364}{365})^n)[/tex]
[tex]-ln\ 3 > n\ ln\ (\frac{364}{365})[/tex]
Apply laws of logarithm
[tex]-ln\ 3 > n\ (ln(364) - ln(365))[/tex]
Multiply through by -1
[tex]ln\ 3 < n\ (ln(365) - ln(364))[/tex]
Solve for n
[tex]\frac{ln\ 3}{(ln(365) - ln(364))} < n[/tex]
Reorder
[tex]n > \frac{ln\ 3}{(ln(365) - ln(364))}[/tex]
[tex]n > \frac{1.09861228867}{(5.89989735358 - 5.89715386764)}[/tex]
[tex]n> \frac{1.09861228867}{0.00274348594}[/tex]
[tex]n > 400.443928891[/tex]
[tex]n > 400[/tex] (approximated)
This implies that n = 401, 402, 403 .....
i.e
[tex]n \geq 401[/tex]
So, at least 401 people has to be invited
