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Chapter 2: Problem 51

A local fraternity is conducting a raffle where 50 tickets are 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?

Short Answer

Expert verified
a) 1/4900, b) 69/4900, c) 207/980, d) 759/980

Step by step solution

01

Define the Problem

We are given 50 raffle tickets, each sold to a different customer. There are 3 prizes to be awarded. We want to calculate the probability of different winning scenarios for 4 specific customers (the organizers). Specifically, we need probabilities for all four organizers winning, exactly two winning, exactly one winning, and none winning.
02

Calculate Probability of the Organizers Winning All Prizes

To win all three prizes, all winners must be among the 4 organizers. The number of ways to choose 3 prize winners out of the 4 organizers is \( \binom{4}{3} = 4 \). The total number of ways to choose 3 winners from 50 customers is \( \binom{50}{3} \). Hence, the probability is \( \frac{4}{\binom{50}{3}} \).
03

Calculate Probability of Exactly Two Organizers Winning

For exactly two organizers to win, we need 2 prize winners from the 4 organizers and 1 from the remaining 46. The number of ways to choose 2 winners from 4 organizers is \( \binom{4}{2} = 6 \), and the number of ways to choose 1 winner from 46 others is \( \binom{46}{1} = 46 \). Thus, the probability is \( \frac{6 \times 46}{\binom{50}{3}} \).
04

Calculate Probability of Exactly One Organizer Winning

To have exactly one organizer winning, choose 1 prize winner from the 4 organizers and 2 winners from the 46 other customers. The number of ways is \( \binom{4}{1} \times \binom{46}{2} \). The probability is \( \frac{4 \times \binom{46}{2}}{\binom{50}{3}} \).
05

Calculate Probability of None of the Organizers Winning

For none of the organizers to win, all 3 prizes must go to the other 46 customers. The number of ways to choose 3 winners from 46 is \( \binom{46}{3} \). Therefore, the probability is \( \frac{\binom{46}{3}}{\binom{50}{3}} \).
06

Calculate Binomial Coefficients and Probabilities

Calculate the required binomial coefficients: \( \binom{50}{3} = 19600 \), \( \binom{46}{3} = 15180 \), and \( \binom{46}{2} = 1035 \). Substitute these values into the probability formulas:- a) \( \frac{4}{19600} = \frac{1}{4900} \)- b) \( \frac{6 \times 46}{19600} = \frac{276}{19600} = \frac{69}{4900} \)- c) \( \frac{4 \times 1035}{19600} = \frac{4140}{19600} = \frac{207}{980} \)- d) \( \frac{15180}{19600} = \frac{759}{980} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a fascinating field of mathematics focused on counting, combinations, and permutations. In simple terms, it helps us understand how to select and arrange objects. This is incredibly useful for probability problems where we need to calculate different possible outcomes. When dealing with raffles, for instance, combinatorics allows us to determine the number of ways to choose winners from a given set of tickets. By understanding combinatorics, we can solve problems involving a finite set of items, such as tickets in a raffle, to identify how many combinations could occur. This forms the bedrock for understanding probability calculations.
Binomial Coefficients
Binomial coefficients are a type of combinatorial tool used for calculating combinations. They help us figure out how many ways we can choose a subset of items from a larger set, where the order doesn't matter. For example, to choose 3 winners from 50 tickets, we use the binomial coefficient \( \binom{50}{3} \). This coefficient is calculated through the formula:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
where \(!\) means factorial, a product of all positive integers up to that number. Binomial coefficients are crucial, especially as they make solving larger combinatorial problems manageable by providing a systematic way to count possibilities.
Raffle Probability Problems
Raffle probability problems involve determining the chance of specific scenarios occurring in a raffle draw. Consider a raffle with 50 tickets and 4 organizers each having a ticket. We might want to know the probability that all organizers, or some, win the prizes. These types of problems can be tackled using combinatorial principles and probability formulas. By calculating the number of favorable outcomes (like the possible winning tickets) and dividing by the total possible outcomes in a raffle, we can derive the probability of these events. This kind of calculation helps in understanding the likelihood of winning in real-life scenarios.
Event Probability Calculation
Event probability calculation is the process of determining how likely an event is to occur. In our raffle example, different scenarios like all, some, or none of the organizers winning are events we might calculate probabilities for. If you want to find the probability of no organizers winning any prize, you compute this as the ratio of favorable outcomes (prizes going to other ticket holders) to the total outcomes.
  • The probability formula used is: \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)
By counting how many ways events can happen and considering all possible scenarios, we derive the likelihood of any given event in the context of the entire problem space.

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Most popular questions from this chapter

Can \(A\) an \(B\) bemutually exclusive if \(P(A)=.4\) and \(P(B)=.7 ?\) If \(P(A)=.4\) and \(P(B)=.3 ?\) Why?

If \(A_{1}, A_{2},\) and \(A_{3}\) are three events and \(P\left(A_{1} \cap A_{2}\right)=P\left(A_{1} \cap A_{3}\right) \neq 0\) but \(P\left(A_{2} \cap A_{3}\right)=0,\) show that \(P\left(\text { at least one } A_{i}\right)=P\left(A_{1}\right)+P\left(A_{2}\right)+P\left(A_{3}\right)-2 P\left(A_{1} \cap A_{2}\right)\)

In a game, a participant is given three attempts to hit a ball. On each try, she either scores a hit, \(H\), or a miss, \(M\). The game requires that the player must alternate which hand she uses in successive attempts. That is, if she makes her first attempt with her right hand, she must use her left hand for the second attempt and her right hand for the third. Her chance of scoring a hit with her right hand is .7 and with her left hand is \(.4 .\) Assume that the results of successive attempts are independent and that she wins the game if she scores at least two hits in a row. If she makes her first attempt with her right hand, what is the probability that she wins the game?

A state auto-inspection station has two inspection teams. Team 1 is lenient and passes all automobiles of a recent vintage; team 2 rejects all autos on a first inspection because their "headlights are not properly adjusted." Four unsuspecting drivers take their autos to the station for inspection on four different days and randomly select one of the two teams. a. If all four cars are new and in excellent condition, what is the probability that three of the four will be rejected? b. What is the probability that all four will pass?

Five firms, \(F_{1}, F_{2}, \ldots, F_{5},\) each offer bids on three separate contracts, \(C_{1}, C_{2},\) and \(C_{3} .\) Any one firm will be awarded at most one contract. The contracts are quite different, so an assignment of \(C_{1}\) to \(F_{1},\) say, is to be distinguished from an assignment of \(C_{2}\) to \(F_{1}\). a. How many sample points are there altogether in this experiment involving assignment of contracts to the firms? (No need to list them all.) b. Under the assumption of equally likely sample points, find the probability that \(F_{3}\) is awarded a contract.
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