At least how many persons are required in a group so that the probability of two persons were born on the same day of the week is $0.5$?
I simplified it so there's only 365 days. Next find the probability that no one is born on the same day of the week then subtract by $1$ to get the solution.
But the answer is wrong, so I'm kind of out of ideas.
Best Answer
Given a group of $n$ people, the sample space of days of the week in which they were born is $\Omega = \left \{ 1, ..., 7 \right \}^n$, which give $\left | \Omega \right |=7^n$.
Let $A$ be the event that at least two of them were born on the same day. Then $P(A)=1-P(A^c)\ $. Now $A^c$ is the set that none of them share a birthday, so you need to pick $n$ different days out of $7$. This can be done in ${7 \choose n}$ ways. Assuming the people are ordered, given a set of $n$ days we have to consider all permutations, hence:
$$P(A^c)={n!{7 \choose n} \over 7^n}$$
Notice that if $n≥8$ then obviously at least two share a birthday, which agrees with our formula since $P(A^c)=0$ for $n>7$. Now use the formula above to compute the probabilities for $2≤n≤7$.