b Use the results from Part A and change only the size of the group of people.
C
c Use the formula found in Part B. To see a table instead of a graph, push 2nd before you push GRAPH.
A
aGroup of 6 People: ≈ 0.04
Group of 10 People: ≈ 0.12
B
b P(n) = 1- _(365)P_n/365^n
C
c 23
Practice makes perfect
a We want to explore a probability problem called the birthday problem. We will begin by finding the probability that at least 2 people share the same birthday in a group of 6 randomly chosen people. In order to solve this problem, we need to consider the complement of the event we are interested in.
A: at least 2 people share the same birthday in a group of 6 people
A: all six people were born on different days
If we find the probability of event A we will be able to obtain the probability of event A, since they are related by the Probability of the Complement of the Event Formula.
The probability of the complement of Event A is P(A) = 1 - P(A).
From now on we will consider Event A. To find P(A), we will use theoretical probability.
P=Favorable Outcomes/Possible OutcomesWe will start by obtaining the number of possible outcomes. Notice the order in which we choose the days is important. We want all 6 people to be born on different days and the birthday date is tied to a particular person.
_n P_r = n!/(n-r)!
We have recalled the formula for the number of permutations of n objects taken r at a time, since permutations take into account the order. In our case we assume there are 365 different birthdays possible, so n = 365. Out of them we choose 6 different days, therefore r= 6. Let's substitute these values into the formula.
We found the number of favorable outcomes. Next, we will look for the number of all possible outcomes. For each person we have the option to choose one day from 365 birthdays possible. That enables us to use the Fundamental Counting Principle.
We substitute 365 for every number of choices of each person's birthday. This is because we are looking for all the possible outcomes, so birthdays can repeat . Let's evaluate the number of possible outcomes.
In order to find the probability of Event A that at least 2 people share the same birthday in a group of 6 people, we have to substitute the obtained probability into the formula for the probability of the complement of event.
To find the probability that at least 2 people share the same birthday in a group of 10 people, we will follow the same way as presented previously. However, the size of the group has to be changed from 6 to 10. We will consider two events.
B: at least 2 people share the same birthday in a group of 10 people
B: all 10 people were born on different days
In order to find P(B), we also assume there are 365 different birthdays possible. This time, to find the number of favorable outcomes we have to substitute 10 for r in the formula for the number of permutations.
The number of possible outcomes this time will be 365^(10), since we just change the number of days from 6 to 10. We are ready to calculate the probability P(B).
In order to find the probability that at least 2 people share the same birthday in a group of 10 people, we will substitute the obtained probability into the formula for the probability of the complement of event.
Notice that when the number of people in the group increases, the probability that at least two people share the same birthday also increases.
b We want to write the formula for the probability P(n) that at least 2 people in a group of n people share the same birthday. Based on results from Part A, we consider two events.
C: at least 2 people share the same birthday in a group of n people
C: all n people were born on different days
In order to find P(C) we will again use theoretical probability.
P=Favorable Outcomes/Possible Outcomes
Let's start by finding the formula for the number of favorable outcomes. This time we want all n people be born on different days. Therefore, we look for the number of permutations of 365 different birthdays possible taken n at a time.
_(365) P_n
We obtained a formula _(365)P_n for the number of favorable outcomes. We will find the number of possible outcomes like in Part A. This time we have to address the fact our group consists of n people. Therefore, there will be 365^n possible outcomes. We are ready to calculate the probability P(C).
P(C) = _(365)P_n/365^n
We aim to find the general formula for the probability that at least 2 people in a group of n people share the same birthday, therefore we need to substitute the obtained probability into the formula for the probability of the complement of event.
Event C corresponds to the event that at least 2 people share the same birthday in a group of n people, therefore we found the desired P(n).
P(n) = 1- _(365)P_n/365^n
c We want to find for what group size the probability that at least 2 people share the same birthday first exceeds 50 %. We will use the table feature on a calculator to make a table of values of the formula from Part B.
P(n) = 1- _(365)P_n/365^n
Therefore, we are interested in solving for n the following inequality.
1 - _(365)P_n/365^n ≥ 0.5
We first have to rewrite the inequality as two functions. One function will be the inequality's left-hand side, and one will be the right-hand side. We also need to replace n with x.
ly=1 - _(365)P_x/365^x y=0.5
To enter the equations on your calculator, push the button Y= and write them in the two first rows.
By pushing 2nd and then GRAPH, we get a table of values for the whole number values of x.
We want to find the x-value that makes the Y_1-column greater than or equal to the Y_2-column. Scroll down until reaching around x=21.
From the table we see that when x=23 the Y_1-column is greater that the Y_2-column for the first time. Therefore, for a group size equal to 23 people the probability that at least 2 people share the same birthday first exceeds 50 %.
