To solve this problem, we'll use the complement rule: we'll find the probability that no two people were born on the same day of the week, and then subtract that from 1 to get the probability that at least two people were born on the same day of the week.
First, let's find the total number of possible ways that the four people could have been born on different days of the week. For the first person, any day of the week is possible. For the second person, there are 6 days of the week left to choose from (since one day has already been taken). For the third person, there are 5 days of the week left to choose from, and for the fourth person, there are 4 days of the week left to choose from. Therefore, the total number of ways that the four people could have been born on different days of the week is:
7 × 6 × 5 × 4 = 840
Now, let's find the total number of possible ways that the four people could have been born on any day of the week (without any restrictions). For each of the four people, there are 7 possible days of the week that they could have been born on. Therefore, the total number of possible ways that the four people could have been born on any day of the week is:
7 × 7 × 7 × 7 = 2401
Therefore, the probability that no two people were born on the same day of the week is:
840/2401 ≈ 0.35
And the probability that at least two people were born on the same day of the week is:
1 - 0.35 = 0.65
So there is a 65% chance that at least two people in the group of four were born on the same day of the week.