Problem 30 - A First Course in Probability

Question

The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that (a) Rebecca and Elise will be paired? (b) Rebecca and Elise will be chosen to represent their schools but will not play cach other? (c) either Rebecca or Elise will be chosen to represent her school?

Step-by-Step Solution

Answer

The short version of the answer to the given question is as follows: a) The probability that Rebecca and Elise will be paired: \[ P_{paired} = \frac{\binom{7}{3} \times \binom{8}{3}}{\binom{8}{4} \times \binom{9}{4} \times 4!}\] b) The probability that Rebecca and Elise will be chosen to represent their schools but will not play each other: \[P_{both\_not\_paired} = \frac{\binom{7}{3} \times \binom{8}{3}}{\binom{8}{4} \times \binom{9}{4}} \times P_{not\_paired}\] c) The probability that either Rebecca or Elise will be chosen to represent her school: \[P_{either\_chosen} = 1 - P_{both\_not\_chosen}\]

Step 1: (Step 1: Calculate Combinations for Representation)

First, let's calculate the possible combinations of 4 players from each school to be chosen to represent their schools. For Rebecca's school, which has 8 players, we can calculate the combinations for 4 members (\(C_1\)) as follows: \[C_1=\binom{8}{4}\] Similarly, for Elise's school with 9 players, we can calculate the combinations for 4 members (\(C_2\)) as follows: \[C_2=\binom{9}{4}\]

Step 2: (Step 2: Calculate Combinations for Pairing)

Now that we have the combinations for both schools, let's calculate the possible pairings for the 4 players from each side (\(P\)). The first player from one school can be paired with any of the 4 players from the other school, the second player can be paired with any of the remaining 3, and so on. So the total number of pairings is: \[P=4 \times 3 \times 2 \times 1 = 4!\]

Step 3: (Step 3: Calculate Probability of Rebecca and Elise Being Paired)

To find the probability of Rebecca and Elise being paired, we need to calculate the combinations of 4 players including Rebecca from her school and 4 players including Elise from her school. For Rebecca's school: \[\binom{7}{3}\] For Elise's school: \[\binom{8}{3}\] The combination of the two sisters being paired with each other is the multiplication of the above two combinations. Now, we need to divide this number by the total number of possible pairings of 4 players from each school to find the probability: \[ P_{paired} = \frac{\binom{7}{3} \times \binom{8}{3}}{\binom{8}{4} \times \binom{9}{4} \times 4!}\]

Step 4: (Step 4: Calculate Probability of Both Sisters Being Chosen but Not Paired)

If both sisters are already chosen and they don't play each other, there are three remaining players in each team who must pair up with each other. \[P_{not\_paired} = \frac{3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\] To calculate the final probability for this case, we multiply the above result with the combinations of both sisters being chosen to represent their schools as in Step 3: \[P_{both\_not\_paired} = \frac{\binom{7}{3} \times \binom{8}{3}}{\binom{8}{4} \times \binom{9}{4}} \times P_{not\_paired}\]

Step 5: (Step 5: Calculate Probability of Either Rebecca or Elise Being Chosen)

To find the probability of either Rebecca or Elise being chosen, we first calculate the probability of neither sister being chosen and then subtract it from 1. Combinations without Rebecca: \[\binom{7}{4}\] Combinations without Elise: \[\binom{8}{4}\] The probability of neither sister being chosen is the multiplication of the above two combinations, divided by the total number of pairings: \[P_{both\_not\_chosen} = \frac{\binom{7}{4} \times \binom{8}{4}}{\binom{8}{4} \times \binom{9}{4} \times 4!}\] Finally, the probability of either Rebecca or Elise being chosen: \[P_{either\_chosen} = 1 - P_{both\_not\_chosen}\]

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arrangements, and combinations. It provides tools to calculate possible ways to choose and arrange objects. Let's break down the ideas for better understanding:

• Combinations: These are selections where the order doesn’t matter. For instance, choosing 4 members from a group of 8 doesn't consider which member is selected first; it's only about which group of 4 is chosen.


• Permutations: These involve arrangements where the order matters. If we needed to arrange members in a line, we would use permutations.

In our chess club scenario, we are primarily dealing with combinations, as the selection of players doesn't depend on the order they are picked. Knowing how to calculate combinations is essential for understanding the probabilities in this problem. For example, the formula for combinations is \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \(n\) is the total number of items to choose from, and \(r\) is the number of items to choose.

Probability of Events

Probability quantifies the likelihood of an event’s occurrence. It ranges between 0 (impossible event) and 1 (certain event). Understanding probabilities involves considering all possible outcomes and those of interest.

• Independent Events: Events that do not affect each other's outcomes.


• Dependent Events: Events where the outcome of one event affects the other.

In our exercise, each pairing of chess players is an independent event. However, being chosen to represent a school affects potential pairing, making it a combinatorial probability challenge. Calculating the probability involves counting the successful outcomes (desired pairings) and dividing by the total possible outcomes. For instance, pairing two specific players (Rebecca and Elise) hinges on the combinations of choosing them from their respective teams and the 4! pairings available.

Chess Club Problem

The problem poses a relatable scenario where we calculate probabilities in a real-world context, involving competitive sports. Here's a summary of how the problem unfolds:

• Two chess clubs, of sizes 8 and 9, respectively, must randomly select 4 members each for a contest.


• Chosen players are paired randomly with each other to play chess games.

It is a perfect example of how probability and combinatorics blend to solve practical problems. This kind of problem helps students visualize mathematical principles in action. Specific tasks include determining the probabilities of both sisters being paired, being in the contest but not paired, or either one being chosen for the contest.

Combinatorial Probability

This concept combines the art of counting with probability calculations. It focuses on determining the likelihood of complex situations by relying on the fundamental rules of counting.

• Combination Calculations: Involves choosing items from groups. In our scenario, we must decide how many ways we can select club members.


• Probability Calculations with Combinations: Once we have all combinations, we calculate the probability by comparing favorable outcomes over the total outcomes.

Let's illustrate the idea with the pairing of Rebecca and Elise. To find their probability of being paired, we calculate the combinations that include both in the representatives, multiply this by pairing possibilities, and divide by total representative combinations and pairing sequences. This illustrates the beauty of combinatorial probability: it methodically breaks down a complex problem into manageable, countable parts.