a Recall the formula for the number of permutations of n objects _nP_n.
B
b Compare the number of favorable outcomes with the outcomes from Part A.
A
a 1/2
B
bProbability: 1/2
Comparison: The probabilities are the same.
Practice makes perfect
a Each day 4 students make presentations of their history project in an order chosen by the teacher. We want to find out the probability that you will be chosen to be the first or second presenter. To find it we will use theoretical probability.
P=Favorable Outcomes/Possible Outcomes
We will start by finding the number of possible outcomes, which is the number of permutations of 4 students presenting their project. We use permutations because the order in which students present is important. Let's recall the formula for the number of permutations of n objects.
_nP_n=n!
Since there are 4 students presenting, in our case n=4. Therefore, the number of possible outcomes is 4!. Next, we will look for the number of favorable outcomes. We want you to be first or the second presenter. Let's first consider the situation where you present first.
Your position is fixed. We need to take into account only the order of the other students.
4-1= 3
There are 3 students left whose order of presenting can be changed. Therefore, we will calculate the number of permutations of 3 other students.
_3P_3= 3!
Now, let's look at the situation when you present second.
Once again we have to consider 3! permutations of 3 other students. To find the number of favorable outcomes, we have to add numbers of permutations obtained in both cases.
The probability that you will be chosen to be the first or second presenter is equal to 12.
b This time we want to find out the probability that you will be chosen to be the second or third presenter of the day. The number of possible outcomes will be again equal to 4!, since there are still 4 students presenting. Let's look for the number of favorable outcomes. We want you to be second or third presenter.
As in Part A, your position is fixed and we consider the number of permutations of 3 other students.
_3P_3= 3!
In the situation when you would present as a third person, once again we have to consider 3! permutations of 3 other students without fixed positions. As in Part A, to find the total number of favorable outcomes we have to add the numbers of combinations from both cases.
3! + 3! = 12
We are ready to calculate the desired probability.
The probability that you will be chosen to be the second or third presenter is equal to 12. Therefore, we conclude that the probabilities obtained in Part A and Part B are the same.
