Problem 60 - Algebra and Trigonometry

Question

A mathematics department consists of ten men and eight women. Six mathematics faculty members are to be selected at random for the curriculum committee. (a) What is the probability that two women and four men are selected? (b) What is the probability that two or fewer women are selected? (c) What is the probability that more than two women are selected?

Step-by-Step Solution

Answer

(a) 0.3166 (b) 0.4365 (c) 0.5635

Step 1: Determine Total Number of Ways to Select Committee

To find the total number of ways to select six members from the 18-member department, we use combinations. The formula for combinations is \( \binom{n}{k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose. Here, \( n = 18 \) and \( k = 6 \).\[\binom{18}{6} = \frac{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 18564\]

Step 2: Calculate Ways to Select Two Women and Four Men

We need the number of ways to pick two women from eight, and four men from ten. The number of combinations for selecting two women from eight is \( \binom{8}{2} \), and for four men from ten is \( \binom{10}{4} \).\[\binom{8}{2} = \frac{8 \cdot 7}{2 \cdot 1} = 28\]\[\binom{10}{4} = \frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} = 210\]

Step 3: Compute Probability of Selecting Two Women and Four Men

Multiply the combinations of selecting two women and four men to find the favorable outcomes.\[28 \times 210 = 5880\]Then, calculate the probability by dividing by the total number of ways to select the committee:\[P(2\, \text{women}, 4\, \text{men}) = \frac{5880}{18564} \approx 0.3166\]

Step 4: Calculate Probability of Selecting Two or Fewer Women

To find the probability of selecting two or fewer women, we sum the probabilities for exactly zero, one, and two women.- **Zero Women:** Select 0 women from 8 and 6 men from 10: \[ \binom{8}{0} \times \binom{10}{6} = 1 \times 210 = 210 \]- **One Woman:** Select 1 woman from 8 and 5 men from 10: \[ \binom{8}{1} \times \binom{10}{5} = 8 \times 252 = 2016 \]- **Two Women:** Already calculated as 5880 ways (see Step 2).Add the favorable outcomes:\[210 + 2016 + 5880 = 8106\]Divide by the total number:\[P(\leq 2\, \text{women}) = \frac{8106}{18564} \approx 0.4365\]

Step 5: Calculate Probability of Selecting More Than Two Women

The probability of selecting more than two women is the complement of selecting two or fewer women.\[P(>2\, \text{women}) = 1 - P(\leq 2\, \text{women}) = 1 - 0.4365 \approx 0.5635\]

Key Concepts

Probability Calculation

Probability calculation is a technique used to measure the likelihood of a specific outcome happening out of a set of possible outcomes. In our problem, we calculate probabilities to find out how likely it is to select a particular combination of women and men for a committee from a mathematics department.

To calculate the probability of an event, you determine the ratio of favorable outcomes to the total number of possible outcomes. For example, the probability of selecting two women and four men involves counting the number of ways to arrange this combination and dividing it by the total ways to form a committee.

In mathematical notation, if an event can occur in "favorable" ways out of "total" ways, the probability \(P\) of that event is:\[ P( ext{event}) = \frac{ ext{favorable outcomes}}{ ext{total outcomes}} \]This foundational concept in probability is vital for understanding more complex scenarios in combinatorics.

Combinations Formula

The combinations formula is a core aspect of solving problems where the order does not matter, such as selecting members of a committee. The formula used is:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \(n\) is the total number of items to choose from, \(k\) is the number of items to choose, and \(!\) denotes factorial, which is the product of all positive integers up to a given number.

For example, in the exercise, the use of \(\binom{18}{6}\) determines all the ways to select any six members from a total of eighteen. Similarly, combinations like \(\binom{8}{2}\) or \(\binom{10}{4}\) find the specific ways to select two women out of eight, or four men out of ten.

This non-permutative method ensures all possible selections are considered, making this formula indispensable in discrete mathematics problems involving selections and groupings.

Committee Selection

Committee selection in combinatorics revolves around choosing a specific number of individuals from a larger group without regard to order. Such problems often involve deciding on how many people from distinct categories (like men and women) will be part of a smaller group (the committee).

For our scenario with the mathematics department, committee selection involves the task of picking six faculty members. Here, a detailed analysis is required to determine combinations, for example, selecting two women and four men, and so forth.

This approach involves calculating each possible desired outcome combination and summing them to get the probability of various scenarios such as having two or fewer women or more than two women in the committee. Utilizing combinations strategically helps in logically determining these potential outcomes.

Discrete Mathematics

Discrete mathematics is a field of study focusing on countable, distinct elements. It is a vital cornerstone of mathematical problems involving finite structures, like the one we're examining.

This branch of mathematics utilizes tools like combinations and permutations to solve problems involving probabilities, set theory, graph theory, and more. Our task of selecting committee members correlates with discrete mathematics as it explores various configurations of groups (women/men combinations) within a finite set (faculty).

Through understanding and applying discrete mathematics, students learn to logically organize, analyze, and derive solutions, making it a practical and robust methodological field. It's this structured approach that aids in clearly presenting and resolving complex combinatoric challenges.