Problem 40 - A First Course in Probability

Question

A town contains 4 people who repair televisions. If 4 sets break down, what is the probability that exactly \(i\) of the repairers are called? Solve the problem for \(i=1,2,3,4 .\) What assumptions are you making?

Step-by-Step Solution

Answer

The probabilities of exactly $i$ repairers being called to fix 4 television sets are: \(P(i=1) = \frac{4}{4^4}\) \(P(i=2) = \frac{\binom{4}{2} * 2^4}{4^4}\) \(P(i=3) = \frac{\binom{4}{3} * (3^4 - 3)}{4^4}\) \(P(i=4) = \frac{4!}{4^4}\) These calculations are based on the assumptions that each television set is repaired by only one repairer and the probability of choosing any repairer is the same for each television set.

Step 1: Understand the assumptions

We are making the following assumptions: 1. Each television set is repaired by only one repairer. 2. The probability of choosing any repairer is the same for each television set.

Step 2: Use the binomial coefficient to find the probabilities

We will calculate the probability of exactly i repairers being called for each case i=1, 2, 3, 4. The total number of possible ways to assign the 4 repair jobs to the 4 repairers is \(4^{4}\), as each of the 4 television sets can be repaired by any of the 4 repairers. For each case, we need to find the number of ways i repairers can be called and divide this by the total number of possible ways.

Step 3: Calculate the probability for i=1

Since there are 4 repairers and we want to find the probability that only one of them repairs all the television sets, the number of ways in this case is 4. Therefore, \[P(i=1) = \frac{4}{4^4}\]

Step 4: Calculate the probability for i=2

For i=2, we need to find the number of ways two repairers can be called to repair the 4 television sets. First, we can choose 2 repairers out of the 4 in \(\binom{4}{2}\) ways. Next, there are 2 television sets each repairer can fix, so the number of ways in which the 4 television sets can be repaired by the 2 chosen repairers is \(2^{4}\). Therefore, \[P(i=2) = \frac{\binom{4}{2} * 2^4}{4^4}\]

Step 5: Calculate the probability for i=3

For i=3, we need to find the number of ways three repairers can be called to repair the 4 television sets. First, we can choose 3 repairers out of the 4 in \(\binom{4}{3}\) ways. Next, for each television set, there are 3 possible repairers who can fix it. The number of ways in which the 4 television sets can be repaired by the 3 chosen repairers without involving the 4th repairer is \((3^4 - 3)\). We subtract 3 because one of the chosen repairers must not repair all 4 television sets. Therefore, \[P(i=3) = \frac{\binom{4}{3} * (3^4 - 3)}{4^4}\]

Step 6: Calculate the probability for i=4

For i=4, we need to find the number of ways all four repairers can be called to repair the 4 television sets. Each repairer should repair exactly 1 television set. This can be done in 4! (4-factorial) ways which represent the number of permutations of 4 elements. Therefore, \[P(i=4) = \frac{4!}{4^4}\] Now, we have calculated the probabilities for each case i=1, 2, 3, 4.

Key Concepts

Binomial Coefficient

The binomial coefficient is a way to count how many ways we can choose a number of items from a larger set without considering the order of selection. In this exercise, it helps us find out how many ways we can select repairers from a total of four individuals. The formula for the binomial coefficient is given by: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(n\) is the total number of items and \(k\) is the number of items to choose. In simpler terms, it helps us understand the different combinations possible, like selecting 2 out of 4 repairers when calculating the probability for \(i=2\). This understanding is crucial as it lays the foundation for determining probabilities based on selection or assignment in probability theory.

Assumptions in Probability

When solving probability problems, certain assumptions are often made to simplify the analysis. These assumptions form the basis for applying probability rules effectively. For this problem, the assumptions are:

• Each television set is repaired by only one repairer, meaning there is no sharing of the task.
• Every repairer has an equal chance of being chosen for any television set, implying fairness and randomness in selections.

These assumptions ensure each event (selection of a repairer for a job) is independent and equally likely, making it easier to calculate the probability for different scenarios. They serve as a critical framework for the application of probability distributions and combinatorics.

Probability Distribution

A probability distribution lists all possible outcomes of a random experiment and their associated probabilities. In our case, the random experiment is assigning repair jobs to repairers, and the outcomes are how many different repairers are called for the repairs.
We calculated the probability for each scenario \(i=1, 2, 3, 4\), forming a probability distribution. For instance, the probability that exactly 2 repairers are called is calculated using the possible combinations of repairers and their assignments. By summing these probabilities, the total should equal 1 if all scenarios are accounted for, illustrating that the distribution covers every possible outcome and is thus complete.

Combinatorial Analysis

Combinatorial analysis is the study of counting, arrangement, and combination possibilities. It is crucial in finding solutions to probability problems that involve multiple possible outcomes.
In the step-by-step solution, combinatorial analysis is applied to determine how many different ways repairers can be assigned to the repair tasks. For example:

• For \(i=2\), \(\binom{4}{2}\) calculates the selection of 2 repairers. Then, considering different assignments for these selections, we use \(2^4\) to reflect all possible ways two repairers can cover four tasks.
• For \(i=4\), \(4!\) permutations give us ways to assign one job per repairer, capturing every possible arrangement of the tasks with four distinct individuals.

Such analysis ensures we don't overlook any possibility in arrangement and allows for the precise calculation of probabilities within the given assumptions.