/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 The City Transit Authority plans... [FREE SOLUTION] | 91Ó°ÊÓ

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The City Transit Authority plans to hire 12 new bus drivers. From a group of 100 qualified applicants, of whom 60 are men and 40 are women, 12 names are to be selected by lot. Suppose that Mary and John Lewis are among the 100 qualified applicants. a. What is the probability that Mary's name will be selected? That both Mary's and John's names will be selected? b. If it is stipulated that an equal number of men and women are to be selected ( 6 men from the group of 60 men and 6 women from the group of 40 women), what is the probability that Mary's name will be selected? That Mary's and John's names will be selected?

Short Answer

Expert verified
a. The probability of Mary being selected without any gender constraint is \(\frac{1}{C(100, 12)}\), and the probability of both Mary and John being selected is \(\frac{C(98, 10)}{C(100, 12)}\). b. With equal gender constraints, the probability of Mary being selected is \(\frac{1}{C(40, 6)}\), and the probability of both Mary and John being selected is \(\frac{C(59, 5)}{C(60, 6)} * \frac{C(39, 5)}{C(40, 6)}\).

Step by step solution

01

Probability of Mary being selected

We will use the definition of probability to find this. It is the ratio of the number of successful outcomes to the total number of possible outcomes. There are 100 qualified applicants, and 12 will be selected. The total number of ways to choose 12 people from 100 is: Total possible outcomes = Combination of 12 from 100: \(C(100, 12)\) There is only one successful outcome, that is Mary being selected among the 12 names. Probability of Mary being selected = \(\frac{1}{C(100, 12)}\)
02

Probability of both Mary and John being selected

Now we have to find the joint probability of Mary and John being selected. There are 98 other applicants besides Mary and John, and we need to choose 10 more individuals from this group since we are considering both Mary and John being selected together. Total possible outcomes = Combination of 10 from 98: \(C(98, 10)\) Probability of both Mary and John being selected = \(\frac{1}{C(100, 12)} * \frac{C(98, 10)}{C(100, 12)}\) Scenario 2: Equal number of men and women
03

Probability of Mary being selected with gender constraint

In this case, we will compute the probability of Mary being selected when 6 men and 6 women are to be chosen. There are 40 women, and 6 will be selected. The total number of ways to choose 6 women from 40 is: Total possible outcomes = Combination of 6 from 40: \(C(40, 6)\) There is only one successful outcome, that is Mary being selected among the 6 names. Probability of Mary being selected with gender constraint = \(\frac{1}{C(40, 6)}\)
04

Probability of both Mary and John being selected with gender constraint

Now we have to find the joint probability of Mary and John being selected with the gender constraint. There are 39 women and 59 men left, besides Mary and John. Since we want equal gender representation, we need to consider that John would be among the 6 men selected. Hence, we need to choose the remaining 5 men from 59 and the remaining 5 women from 39. Total possible outcomes = Combination of 5 from 59: \(C(59, 5)\) and Combination of 5 from 39: \(C(39, 5)\) Probability of both Mary and John being selected with gender constraint = \(\frac{1}{C(40, 6)} * \frac{C(59, 5)}{C(60, 6)} * \frac{C(39, 5)}{C(40, 6)}\) In conclusion, we found the probabilities of both individual events and joint events in each of the two scenarios.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arranging, and finding patterns. It's all about figuring out the number of possible configurations or combinations in a given set.
  • Basic Counting Principle: Counts the number of outcomes in sequences of events.
  • Permutations: Arrangements of objects in a specific order.
  • Combinations: Selection of objects without regard to order, used as in our exercise when choosing names from a list of applicants.
In our problem, we're dealing with combinations as we select 12 names from a group of 100 applicants. We're not concerned with the order in which names are drawn, just the selection of names itself. Using combinations, the total number of ways to select 12 people from 100 is denoted by the binomial coefficient, which is calculated using:\[C(n, k) = \frac{n!}{k!(n-k)!}\]Where \( n \) is the total number of applicants (100), and \( k \) is the number of selections (12). By understanding these concepts, one can efficiently determine probabilities in scenarios with large choices.
Conditional Probability
Conditional probability looks at the likelihood of an event occurring given that another event has already occurred. This helps us narrow down the field when certain conditions are in place.When analyzing the conditional probability in scenarios like our exercise, you're often dealing with the relationship between two events. For example, you might be asked to find the probability of Mary being chosen when certain requirements, like gender balance, are imposed.To calculate conditional probability, use the formula:\[P(A | B) = \frac{P(A \cap B)}{P(B)}\]This expression means you're finding the probability of event \( A \) happening given that event \( B \) is true. In the context of our exercise:
  • Calculate the probability of selecting Mary's name assuming that half the chosen 12 must be women.
  • It changes the total number of outcomes to only include combinations that maintain this balance.
We adapt our understanding of chances in situations where conditions or restrictions apply, enhancing our ability to assess different real-world scenarios.
Binomial Coefficient
A binomial coefficient is a specific kind of combination number, which is used extensively in combinatorics to count the number of ways to choose "k" objects from "n" objects without considering the order. Often represented as \( C(n, k) \) or \( \binom{n}{k} \), it forms the backbone of binomial probability calculations.Mathematically, it's expressed and calculated as:\[C(n, k) = \frac{n!}{k!(n-k)!}\]where \( n! \) ("n factorial") is the product of all positive integers up to \( n \). Factorials grow rapidly, but they can be easily managed with modern calculators or computer programs.In our example, we use the binomial coefficient to calculate how many different ways we can select 12 names from 100 applications. Here,
  • \( n = 100 \): total pool of applicants.
  • \( k = 12 \): spots available for selection.
Understanding this coefficient is crucial to working with probability problems involving selections, as it effectively quantifies the possible configurations you might encounter.

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