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:

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

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