Question

How many people have to be in a room in order that the probability that at least two of them celebrate their birthday on the same day is at least 0.06? (Ignore leap years, and assume that all outcomes are equally likely.)

Answer 1

At least 8 people are needed in the room

Explanation:

The probability of n people celebrating their birthday in different days is

1*364/365*363/365*....*(365-(n-1))/365

(for the first person any of the 365 days is suitable, for the second person only 364 persons is suitable, for the third only 364 and so on)

At least two people celebratig their birthday on the same day is the complementary event, thus its probability is

1- 364/365*....*(365-n+1)/365

Lets compute the probabilities for each value of n:

• for n = 2: 1-364/365 = 0.00273
• for n = 3: 1-364/365*363/365 = 0.008
• for n = 4: 1-364/365*363/365*362/365 = 0.016
• for n = 5: 1-364/365*363/365*362/365*361/365 = 0.027
• for n = 6: 1-364/365*363/365*362/365*361/365*360/365 = 0.04
• for n = 7: 1-364/365*363/365*362/365*361/365*360/365*359/365 = 0.056
• for n = 8: 1-365/365*363/365*362/365*361/365*360/365*359/365*358/365 = 0.074 > 0.06

We need at least 8 people in the room.