It is of course unrealistic that the chance of a person born on a Friday is $\frac{1}{3}$, but so be it. We are also told that for all other days of the week the chance is the same, so that would be
$$\frac{1-\frac{1}{3}}{6}=\frac{1}{9}$$ (e.g. the chance of being born on a Wednesday is $\frac{1}{9}$)
Now, there are various ways to have 2 people being born on the same day, and the other two on 2 other days:
$A$. Two born on a Friday, and the other two on two different days other than Friday
$B$. Two born on the same day but other than a Friday, one born on Friday, and the last on a different day yet.
$C$. Two born on the same day but other than a Friday, and the other two on two different days other than Friday
$$P(A) = {4 \choose 2} \cdot \big( \frac{1}{3} \big)^2\cdot \frac{6}{9} \cdot \frac{5}{9} = \frac{180}{729}$$
(the $4 \choose 2$ is the number of ways to pick the two people born on a Friday)
$$P(B) = {4 \choose 2} \cdot 6 \cdot \big( \frac{1}{9} \big)^2\cdot 2 \cdot \frac{1}{3} \cdot \frac{5}{9} = \frac{120}{729}$$
(the extra $2$ term is to distinguish the two people: one born on Friday, the other not. The $6$ is to pick one of the 6 days that the first two are born on)
$$P(C) = {4 \choose 2} \cdot 6 \cdot \big( \frac{1}{9} \big)^2 \cdot \frac{5}{9} \cdot \frac{4}{9} = \frac{80}{729}$$
Hence, the chance that 2 people being born on the same day, and the other two on 2 other days is:
$$P(A)+P(B)+P(C)=\frac{380}{729} \approx 0.52$$
Best Answer
Your calculation is based on the assumption the events are mutually exclusive. However your calculation of P(2 people) is actually P(2 or more), while P(3 people) is actually P(3 or more). For example P(exactly 3 people) should be $\frac{364}{365^3}\binom{4}{3}$.