Problem 5
Question
A group consists of four men and five women. Three people are selected to attend a conference. a. In how many ways can three people be selected from this group of nine? b. In how many ways can three women be selected from the five women? c. Find the probability that the selected group will consist of all women.
Step-by-Step Solution
Verified Answer
a) There are 84 ways to select any three people from the group of nine people. b) There are 10 ways to select three women from the five. c) The probability that all the members of the selected group are women is 0.119.
1Step 1: Total Ways of Selection
For part (a), we need to choose any three people out of total nine. In combinations, order doesn't matter. The general formula for combinations is given by \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of items, k is the number of items to choose, and '!' stands for factorial. Applying this formula, we get \(C(9, 3) = \frac{9!}{3!(9-3)!}\)
2Step 2: Selection of Women
For part (b), we should only consider women for the selection pool. With now only five women, the question, again, becomes a matter of calculating a combination. Substituting into our formula, we get \(C(5, 3) = \frac{5!}{3!(5-3)!}\)
3Step 3: Probability Calculation
For part (c), we need to divide the number of ways to choose three women by the total number of ways to choose three people. If A represents the event of '3 women selected' and S represents the 'total ways of selection', then Probability (P) is given by \(P(A/S) = \frac{C(5, 3)}{C(9, 3)}\)
Key Concepts
ProbabilityFactorialCombinationsSelection problems
Probability
Probability is a way of measuring the likelihood of an event happening. It is expressed as a number between 0 and 1, where 0 means the event will not happen, and 1 means it definitely will. When dealing with selection problems, like choosing from a group, we use probability to predict the chances of a particular selection. For example, if you want to find the probability of selecting all women from a group, you compare the number of ways to select the group of women to the total number of selection possibilities.
- If there are fewer ways to make the selection, the probability will be lower.
- The general formula for probability is \[ P(A) = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}} \]
Factorial
Factorials are a crucial concept in combinatorics. They are used to simplify the calculation of permutations and combinations. A factorial, represented by an exclamation mark (!), is the product of all positive integers up to a given number. For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Factorials grow very fast as the number increases, which is why they're useful in calculating large numbers of possible arrangements or selections.
Factorials grow very fast as the number increases, which is why they're useful in calculating large numbers of possible arrangements or selections.
- Factorials help us organize and count the number of ways to arrange or select items from a set.
- They are an essential part of the formula used in combinations and permutations.
Combinations
Combinations are selections made from a larger group where the order does not matter. This is different from permutations, where the order does matter. In our context, combinations help us determine the number of ways to select a subset of people from the group. The formula we use to calculate combinations is: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items and \( k \) is the number of items to choose.
- Combinations are used when you want to know how many different groups can be formed.
- They are widely used in probability calculations to evaluate possible selection outcomes.
Selection problems
Selection problems involve determining the number of ways to select items or people from a larger group. These are common in probability and combinatorics, where order typically doesn't matter. Problems like choosing committees or picking a team fit this category.
- Often, these problems require calculating combinations or permutations.
- Key to solving them is identifying whether order matters and choosing the right formula.
Other exercises in this chapter
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