A raffle is being conducted with 50 tickets to be sold - one per customer.
There are three prizes to be awarded.
If the four organizers of the raffle each buy one ticket, what is the probability that the four organizers win:
a) all of the prizes?
b) exactly two of the prizes
c) exactly one of the prizes
d) none of the prizes
This is at the end of a chapter on counting combinations/permutations
I know the fundamental formulas and their derivations really well. This problem demands a more complicated application that I'm struggling with...
all of the prizes:
4/50*3/49*2/48*1/47
Two of the prizes:
4/50*3/49*46/48*45/47
See the pattern?
To solve this problem, we'll need to use the concepts of combinations and permutations.
First, let's calculate the total number of combinations for the tickets. Since there are 50 tickets and each customer can purchase only one, the total number of combinations is given by:
Total Combinations = 50!
Where "!" denotes the factorial of a number.
a) Probability that the four organizers win all of the prizes:
In this case, each organizer needs to win one of the three prizes. Since there are three prizes and four organizers, the total number of combinations for the four organizers to win all of the prizes is given by:
Combinations = 3!
To find the probability, divide the number of combinations by the total number of combinations:
Probability = Combinations / Total Combinations
b) Probability that the four organizers win exactly two of the prizes:
To calculate this probability, we need to determine the number of ways to choose two out of the three prizes for the organizers and multiply it by the number of ways to assign one prize to each of the chosen organizers.
Number of combinations to choose 2 out of 3 prizes = C(3, 2)
Number of ways to assign one prize to each of the chosen organizers = 2!
Total combinations for exactly two organizers to win = C(3, 2) * 2!
Now, divide the number of combinations by the total number of combinations to get the probability.
c) Probability that the four organizers win exactly one of the prizes:
Following a similar logic as in part b, we need to choose one prize for the organizers and assign one organizer to each of the chosen prizes.
Number of combinations to choose 1 out of 3 prizes = C(3, 1)
Number of ways to assign one organizer to each of the chosen prizes = 1!
Total combinations for exactly one organizer to win = C(3, 1) * 1!
Again, divide the number of combinations by the total number of combinations to get the probability.
d) Probability that the four organizers win none of the prizes:
In this case, none of the organizers can win any of the prizes. The number of combinations for this scenario is simply 0.
Now that we have the formulas, you can substitute the values into the expressions to calculate the probabilities.