Problem 41 - Algebra and Trigonometry

Question

A committee of five is chosen randomly from a group of six males and eight females. What is the probability that the committee includes either all males or all females?

Step-by-Step Solution

Answer

The probability that the committee includes either all males or all females is \( \frac{31}{1001} \).

Step 1: Identify Total Number of People

First, we need to find the total number of people in the group. We have 6 males and 8 females, so the total is 6 + 8 = 14 people.

Step 2: Use Combinatorial Formula for Total Committees

The total number of ways to choose 5 people out of 14 is given by the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). Here, \( n = 14 \) and \( r = 5 \). So, the total combinations are \( \binom{14}{5} \).

Step 3: Calculate \( \binom{14}{5} \)

Calculate \( \binom{14}{5} = \frac{14!}{5!(14-5)!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002 \).

Step 4: Calculate Number of All Male Committees

To calculate the number of ways to choose all males, we need 5 males, which is not possible as there are only 6 males. Therefore, \( \binom{6}{5} = 6 \).

Step 5: Calculate Number of All Female Committees

To calculate the number of ways to choose all females, we need 5 females out of 8. So, we use \( \binom{8}{5} \).

Step 6: Calculate \( \binom{8}{5} \)

Compute \( \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \).

Step 7: Add Cases for All Male and All Female Committees

Since these events (all males, all females) are mutually exclusive, add the number of ways each event can happen: \( 6 + 56 = 62 \).

Step 8: Calculate the Probability

The probability that the committee is either all males or all females is the number of favorable outcomes divided by total outcomes. So, the probability is \( \frac{62}{2002} \).

Step 9: Simplify the Probability Fraction

Simplify \( \frac{62}{2002} \). The greatest common divisor (GCD) of 62 and 2002 is 2, so the simplified fraction is \( \frac{31}{1001} \).

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination. In our example, we solve the problem by determining how many ways we can choose a committee from a group. This involves counting and arranging different selections that can be made. It's like finding all possible variations of a set. Combinatorics uses specific formulas that help us calculate the number of possible ways events can occur.

For instance, when calculating how many committees of five can be picked from 14 people, we apply combinatorial principles. The formula for combinations is denoted as \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). This formula helps find out how many ways there are to pick \( r \) items from \( n \) items without caring about the order.

• "\( \binom{n}{r} \)" denotes "n choose r" reflection, which specifies selection without replacement.
• "\( n! \)" or n factorial, multiplying all positive integers up to \( n \).

Combinatorics simplifies the decision-making by providing mathematical methods for tallying outcomes effectively.

Permutations and Combinations

Understanding permutations and combinations is crucial when calculating probabilities in combinatorics. Permutations consider arrangements where order does matter, whereas combinations focus on selections where order does not matter. In our problem, we deal with combinations.

When selecting a committee, the order of selection doesn't matter. For example, choosing person A and then person B is the same as choosing person B first, then A. Thus, combinations are the most suitable approach.

• Permutations formula: \( nPr = \frac{n!}{(n-r)!} \)
• Combinations formula: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

In our exercise, we used combinations to determine the number of ways to form committees: One with 5 males, even if impractical given the limited number of males, and another with 5 females from 8 potential candidates.
This separation illustrates how permutations and combinations can effectively distinguish problems related to arrangement versus selection.

Mutually Exclusive Events

Mutually exclusive events are events that cannot happen at the same time. In our example, the event "a committee of all males" and "a committee of all females" cannot occur simultaneously. Understanding this concept is key to solving the exercise.

Because these events are mutually exclusive, we don't double-count any outcome. Instead, we can add the separate probabilities. If event A and event B cannot happen together, the probability of either A or B happening is simply the sum of their individual probabilities.

• For example: If the probability of event A is 0.3 and event B is 0.4, and they are mutually exclusive, then \( P(A \cup B) = P(A) + P(B) = 0.7 \).

In our problem, this means the probability of forming an all-male group plus the probability of an all-female group can be combined:

• The number of all-male selections (6) plus all-female selections (56) yields 62 out of possible total 2002 combinations.

Simplifying the calculation follows basic arithmetic principles to find a reduced probability of forming such committees. So, the probability is calculated as having 62 favorable ways of forming an all-male or all-female committee out of 2002 total combinations.