What is the probability that in a company of 500 people, only two persons will have birthdays on New Year’s Day? I feel the answer is
=$\frac{1}{365}\cdot\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{364}{365}\cdots\frac{364}{365}$ =$$\frac{1}{365^{2}}.\frac{364^{498}}{365^{498}}$$
And Not= $$\binom{500}2\cdot\frac{1}{365^{2}}\cdot\frac{364^{498}}{365^{498}}$$
Basically in the second answer they have first selected 2 out of 500 people...which I am not sure is necessary here or not... Coz what I gave a thought when I saw the problem ..that only 2 people means ..there are some predetermined or fixed 2 people and so I am not supposed to select first 2 out of 500 before I go in solving this question... (Plz tell was my thought actually making any sense..??)
This is a problem related to the Binomial distribution. Maybe it helps considering the smaller case of having two out of four people with a birthday on New Year's day. Let's define the events $A$, $B$, $C$ and $D$ as the events in which the first, second, third and fourth person has their birthday on $1$ January. Then, the following cases are all valid:
$$A, B, C^\mathsf{c}, D^\mathsf{c}$$ $$A, B^\mathsf{c}, C, D^\mathsf{c}$$ $$A, B^\mathsf{c}, C^\mathsf{c}, D$$ $$A^\mathsf{c}, B, C, D^\mathsf{c}$$ $$A^\mathsf{c}, B, C^\mathsf{c}, D$$ $$A^\mathsf{c}, B^\mathsf{c}, C, D$$
The number of valid cases equals ${4 \choose 2} = 6$, i.e., the number of ways in which we can select two people out of four. The probability of each of these cases to appear, equals:
$$\frac{1}{365}\frac{1}{365}\frac{364}{365}\frac{364}{365} = \frac{364^2}{365^4}$$
Therefore, the probability of having exactly two out of four people whose birthday is on a specific day, equals:
$${4 \choose 2} \frac{364^2}{365^4} \approx 0.0000448$$
In general, if an event occurs with probability $p$, the probability of having $k$ out of $n$ occurrences equals:
$${n \choose k} p^k (1-p)^{n-k}$$