/*! 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 92 An employee of the records offic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An employee of the records office at a certain university currently has ten forms on his desk awaiting processing. Six of these are withdrawal petitions and the other four are course substitution requests. a. If he randomly selects six of these forms to give to a subordinate, what is the probability that only one of the two types of forms remains on his desk? b. Suppose he has time to process only four of these forms before leaving for the day. If these four are randomly selected one by one, what is the probability that each succeeding form is of a different type from its predecessor?

Short Answer

Expert verified
a. Probability is \( \frac{1}{105} \). b. Probability is \( \frac{1}{7} \).

Step by step solution

01

Understanding the Problem

We have a total of ten forms on the desk: six are withdrawal petitions (W) and four are course substitution requests (S). We need to calculate two probabilities for two different scenarios.
02

Calculate Total Combinations for Part (a)

The total number of ways to select six forms out of ten is given by the combination formula \( \binom{10}{6} \). Calculate \( \binom{10}{6} = 210 \). This represents all possible ways to select any six forms from the ten forms available.
03

Calculate Combinations for Part (a) - Single Type Left

For either only withdrawal petitions (W) or course substitution requests (S) to be left on the desk, the subordinate can take all four of S (in one scenario) or all six of W (in another scenario).\- Scenario W: \( \binom{6}{6} = 1 \) and \( \binom{4}{0} = 1 \) because all 6 W are selected.\- Scenario S: \( \binom{6}{0} = 1 \) and \( \binom{4}{4} = 1 \) because all 4 S are selected.\The number of favorable ways is the sum of these two scenarios: \( 1 + 1 = 2 \).
04

Calculate Probability for Part (a)

The probability that only one type of form remains on his desk is the number of favorable outcomes divided by the total outcomes: \( \frac{2}{210} = \frac{1}{105} \).
05

Understand the Scenario for Part (b)

We now consider him processing four forms, and they must be selected such that each form is a different type than the one before it. There are two valid sequences possible: WSWS and SWSW.
06

Calculate Probability for Sequence WSWS

Let's find the number of favorable outcomes for the sequence where the pattern is WSWS:\- Choose W first: \( 6 \) choices (since there are 6 W forms initially), \- then S: \( 4 \) choices (4 S forms remaining), \- then W: \( 5 \) choices (one W already taken, so 5 left), \- and finally S: \( 3 \) choices (one S already taken, so 3 left). \The number of satisfied sequences is \( 6 \times 4 \times 5 \times 3 = 360 \).
07

Calculate Probability for Sequence SWSW

Do the same calculation for the sequence starting with SWSW: \- Choose S first: \( 4 \) choices,- then W: \( 6 \) choices,- then S: \( 3 \) choices (one S is already selected),- and finally W: \( 5 \) choices.The result is the same number of satisfied sequences: \( 4 \times 6 \times 3 \times 5 = 360 \).
08

Calculate Total Favorable Outcomes for Part (b)

Add the results from sequences WSWS and SWSW: \( 360 + 360 = 720 \).
09

Calculate Total Possible Outcomes for Part (b)

The total number of ways to choose any 4 forms out of 10 is \( \binom{10}{4} = 210 \). Multiply by all permutations within those 4: \( 4! = 24 \). Thus, total outcomes \( 210 \times 24 = 5040 \).
10

Calculate Probability for Part (b)

Divide the number of favorable outcomes by the total possible outcomes: \( \frac{720}{5040} = \frac{1}{7} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Combination Formula
In combinatorial probability, understanding the combination formula is key to solving problems where order does not matter. The combination formula helps determine how many ways you can choose items from a larger set. It's denoted as \( \binom{n}{r} \), where \( n \) is the total number of items, and \( r \) is the number of items to be chosen. The formula for combinations is \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] This formula calculates how many possible sets you can form. For example, in part (a) of our exercise, the employee needs to choose 6 forms out of 10. Applying the combination formula: \[ \binom{10}{6} = \frac{10!}{6!(10-6)!} = 210 \] This represents all possible sets of 6 forms.
  • Factorials (\(!\)) are repeated product sequences of integers less than or equal to \( n \).
  • The order of selection is irrelevant, making it a combination, not a permutation.
Using combinations allows you to focus purely on the selection process without considering order.
Probability Calculation
Probability calculations in the given exercise rely on understanding favorable outcomes versus total possible outcomes. In probability, each experiment consists of a sample space with possible outcomes. Probability is calculated using the formula: \[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \] For part (a), the issue is determining the probability that only one type of form is left.
  • Total combinations possible: \( \binom{10}{6} = 210 \).
  • Favorable scenarios: All 6 withdrawal petitions selected (\( \binom{6}{6} \)) or all 4 substitutions selected (\( \binom{4}{4} \)).
    • Each gives only 1 way, resulting in 2 favorable outcomes in total.
  • Calculation: \( P = \frac{2}{210} = \frac{1}{105} \).
In part (b), we focus on selecting forms so each successive is different. We found valid sequences: WSWS and SWSW. Each has 360 favorable ways, totaling 720 favorable outcomes versus 5040 total sequences possible. Thus, \[ P = \frac{720}{5040} = \frac{1}{7} \] Effectively, probability calculations turn complex scenarios into manageable numbers.
Problem Solving Steps
Approaching combinatorial problems requires a systematic methodology. Here is a step-by-step approach:
1. **Understand the Problem:** Break down what is being asked. Identify known values and variables.

2. **Determine Total Combinations:** Use the combination formula to identify the total number of outcomes. For the exercise, calculating \( \binom{10}{6} \) gives the total selections.

3. **Identify Favorable Cases:** Look for specific outcomes that meet the problem's criteria. For instance, isolating all one form type or alternating sequences.

4. **Calculate Probability:** Divide the number of favorable outcomes by the total possible outcomes, applying the fundamental probability formula.

5. **Verify Results:** Cross-check calculations for accuracy. Ensure the logic and math consistently reflect the problem setup.

This structured process fosters clear thinking and accuracy in dealing with probability and combinations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\left(\begin{array}{c}n \\ n-k\end{array}\right)\). Give an interpretation involving subsets.

As of April 2006, roughly 50 million .com web domain names were registered (e.g., yahoo.com). a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits as well as letters are permitted as characters? [Note: A character length of three or more is now mandated.] b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? [Note: All are currently taken.] c. Answer the questions posed in (b) for four-character sequences. d. As of April \(2006,97,786\) of the four-character sequences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?

1 and #2. If one pump fails, the system will still operate. However, because of the added strain, the remaining pump is now mo… # A system consists of two identical pumps, #1 and #2. If one pump fails, the system will still operate. However, because of the added strain, the remaining pump is now more likely to fail than was originally the case. That is, \(r=P(\\# 2\) fails \(\\# 1\) fails \()>P\) (#2 fails) \(=q\). If at least one pump fails by the end of the pump design life in \(7 \%\) of all systems and both pumps fail during that period in only \(1 \%\), what is the probability that pump #1 will fail during the pump design life?

A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles. Let \(A\) denote the event that the New York flight is full and define events \(B\) and \(C\) analogously for the other two flights. Suppose \(P(A)=.6, P(B)=.5, P(C)=.4\) and the three events are independent. What is the probability that a. All three flights are full? That at least one flight is not full? b. Only the New York flight is full? That exactly one of the three flights is full?

1,30 \%\( of the time on airline \)\\#… # A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; \(50 \%\) of the time she travels on airline \(\\# 1,30 \%\) of the time on airline \(\\# 2\), and the remaining \(20 \%\) of the time on airline \(\\# 3\). For airline \(\\# 1\), flights are late into D.C. \(30 \%\) of the time and late into L.A. \(10 \%\) of the time. For airline \(\\# 2\), these percentages are \(25 \%\) and \(20 \%\), whereas for airline \(\\# 3\) the percentages are \(40 \%\) and \(25 \%\). If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines \(\\# 1, \\# 2\), and #3?Assume that the chance of a late arrival in L.A. is unaffected by what happens on the flight to D.C. [Hint: From the tip of each first-generation branch on a tree diagram, draw three second-generation branches labeled, respectively, 0 late, 1 late, and 2 late.]

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.