Question

Need a Python code that will run in spyder Anacoda for the following Mathmatical problem.

Q1. Consider an experiment that consists of recording the birthday for each of 20 randomly selected persons. Ignoring leap years, we assume that there are only 365 possible distinct birthdays. Furthermore, we assume that each of the possible sets of birthdays is equi-probable

(1) What is the probability that each person in the 20 has a different birthday?

(2) Find the minimum number of persons such that the probability of two or more people have a same birthday is at least 50%.

(3) Find the minimum number of persons such that the probability of two or more people have a same birthday is at least 95%.

Answer 1

#CODE

""" Q1 Answer
Approach:
   Each person should have a different birthday.
   For 1 person -> probability = 365/365 = 1
   For 2 persons -> probability = 365/365 * (364/365) #As second person has 364 different possiblites excpet one for the first person
   For 3 person -> probability = (365/365) * (364/365) * (363/365)
   ....
"""
n=20
P = 1
for i in range(20):
   P = P * ((365-i)/365)

print("1 Answer:\nThe probability that each person in the 20 has a different birthday : ",P)

def personsRequired(probability):
   """
       We need no.of persons required with the same birthdate with
       given probability. Let's approach this in reverse way that
       are we find people required for 1-probability i.e, people required
       for having different birthdays and then add 1 to them to get two people
       birthday with a given probability.
   """
   negativeProb = 1-probability
   start = 365
   i = 0
   persons = 0
   P = 1
   while(P > negativeProb):
       P = P * ((start-i)/365)
       i+=1
       persons+=1
   return persons

print("2 Answer:\nminimum number of persons such that the probability of two or more people have a same birthday is at least 50% :",personsRequired(0.5))
print("3 Answer:\nminimum number of persons such that the probability of two or more people have a same birthday is at least 95% :",personsRequired(0.95))

#OUTPUT