What is the probability that at least 2 people in a random group of 4 people have a birthday on the same day of the week? Answer:
The opposite event is that 4 people have
birthdays on different days of the week.
Its probability=(7*6*5*4)/(7^4)=120/343
Therefore, P=1-120/343
sum(dbinom(2:4,4,1/7)) = 0.1003748
To calculate the probability that at least 2 people in a random group of 4 people have a birthday on the same day of the week, we can use the concept of complementary probability.
First, we need to find the probability that none of the 4 people have a birthday on the same day of the week.
There are 7 days in a week, so the first person can have their birthday on any day of the week, giving us a probability of 7/7.
For the second person, there are now 6 remaining days of the week that they can have their birthday on, giving us a probability of 6/7.
Similarly, the third person has 5 remaining days of the week to choose from, giving us a probability of 5/7.
Finally, the fourth person has 4 remaining days of the week to choose from, with a probability of 4/7.
To find the probability that none of the 4 people have a birthday on the same day of the week, we multiply these probabilities together:
(7/7) * (6/7) * (5/7) * (4/7) = 120/2401
Next, we can calculate the probability that at least 2 people have a birthday on the same day of the week by subtracting the probability of none of them having the same birthday:
1 - (120/2401) = 2281/2401
Therefore, the probability that at least 2 people in a random group of 4 people have a birthday on the same day of the week is 2281/2401 or approximately 0.95 or 95%.