Problem 422

Determine the probability that at least 2 people in a room of 4 people share the same birthday, ignoring leap years and assuming each birthday is equally likely, by answering the following question:

(a)    Compute the probability that 4 people have different birthdays.

(b)   The complement of “ 4 people have different birthdays” is “at least 2 share a birthday”. Use this information to compute the probability that at least 2 people out of 4 share the same birthday.

 

Solution

 

(a)    The probability that 4 people have different birthday can be found by applying the general multiplication rule many times.

P(4 different birthdays) = P(2nd different) * P(3rd different)*…….*P(4th different)

Start with the first person. There are 365 possible birthdays for this person. Therefore, the probability that the first person has a birthday is \frac{365}{365} = 1.

Multiply this by the probability that the second person’s birthday is different. Since one day is taken by the first person, there are 364 possible birthdays for the second person.

P(2 different birthdays) = \frac{365}{365} * \frac{364}{365}

Multiply this by the probability that the third person’s birthday is different. Now there are only 363 days available.

P(3 different birthdays) = \frac{365}{365} * \frac{364}{365}*\frac{363}{365}

Notice that the numerator decreases by 1 for each probability. Continue this process for all 4 people.

P(4 different birthdays) = \frac{365}{365}*\frac{364}{365}*\frac{363}{365}*\frac{362}{365}

Multiple all the probabilities to find the probability that 4 people have different birthdays.

P(4 different birthdays) \overall 0.9836

 

(b)   Now find the complement of this probability.

P(at least 2 share a birthday) = 1 – P(4 different birthdays)

\overall 1 – 0.9836

\overall 0.0164

 

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