Problem 46 - A First Course in Probability

Question

How many people have to be in a room in order that the probability that at least two of them celebrate their birthday in the same month is at least \(\frac{1}{2} ?\) Assume that all possible monthly outcomes are equally likely.

Step-by-Step Solution

Answer

The minimum number of people that need to be in a room in order for the probability that at least two of them celebrate their birthdays in the same month to be at least \(\frac{1}{2}\) is \(n = 5\).

Step 1: Define the probability space

The total number of equally likely ways to distribute the birthdays of 'n' people among 12 months is \(12^n\). Each person can have their birthday in one month, and since there are 12 choices for each of the 'n' people, there are \(12^n\) possible combinations.

Step 2: Calculate the complementary probability

We are looking for the probability that at least two people have their birthdays in the same month, so using the complementary rule, we will find the probability that they all have their birthdays in different months. The number of ways they can have their birthdays in different months is \(12 \times 11 \times 10 \times ... \times (12-n+1)\), as the first person has 12 month options, the second person has 11 remaining months to choose from, and so on till the nth person.

Step 3: Finding the desired probability

To find the probability that all the people in the room have their birthdays in different months, we divide the number of ways to have distinct birthdays by the total number of equally likely ways to distribute the birthdays. This gives us \(\frac{12 \times 11 \times 10 \times ... \times (12-n+1)}{12^n}\).

Step 4: Establishing the inequality

We want to find the smallest positive integer 'n' such that more than half the time, at least two people share a birthday month. Equivalently, we want to find the smallest 'n' satisfying the inequality: \( 1 - \frac{12 \times 11 \times 10 \times ... \times (12-n+1)}{12^n} \geq \frac{1}{2} \)

Step 5: Solving the inequality

We can simplify the inequality: \(\frac{12 \times 11 \times 10 \times ... \times (12-n+1)}{12^n} \leq \frac{1}{2}\) Now, we can start plugging in different values for 'n' and sampling probabilities until the inequality holds. n = 1: \( \frac{12}{12} = 1 \) (Not satisfactory) n = 2: \( \frac{12 \times 11}{12^2} = \frac{11}{12} \) (Not satisfactory) n = 3: \( \frac{12 \times 11 \times 10}{12^3} = \frac{11 \times 10}{12 \times 12} = 0.763 \) (Not satisfactory) n = 4: \( \frac{12 \times 11 \times 10 \times 9}{12^4} = \frac{11 \times 10 \times 9}{12 \times 12 \times 12}= 0.592 \) (Not satisfactory) n = 5: \( \frac{12 \times 11 \times 10 \times 9 \times 8}{12^5} = \frac{11 \times 10 \times 9 \times 8}{12 \times 12 \times 12 \times 12}= 0.397 \) (Satisfactory) The smallest positive integer 'n' that satisfies the inequality is 5. Thus, there must be at least 5 people in the room in order for the probability that at least two of them celebrate their birthdays in the same month to be at least \(\frac{1}{2}\).