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Chapter 2: Problem 92

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: \( \frac{720}{5040} = \frac{1}{7} \).

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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.

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Most popular questions from this chapter

An ATM personal identification number (PIN) consists of four digits, each a \(0,1,2, \ldots 8\), or 9 , in succession. a. How many different possible PINs are there if there are no restrictions on the choice of digits? b. According to a representative at the author's local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited: (i) all four digits identical (ii) sequences of consecutive ascending or descending digits, such as 6543 (iii) any sequence starting with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the probability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)? c. Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the \(2^{\text {nd }}\) and \(3^{\text {rd }}\) digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the individual knows about the restrictions described in (b) so selects only from the legitimate possibilities). What is the probability that the individual gains access to the account? d. Recalculate the probability in (c) if the first and last digits are 1 and 1 , respectively.

In October, 1994, a flaw in a certain Pentium chip installed in computers was discovered that could result in a wrong answer when performing a division. The manufacturer initially claimed that the chance of any particular division being incorrect was only 1 in 9 billion, so that it would take thousands of years before a typical user encountered a mistake. However, statisticians are not typical users; some modern statistical techniques are so computationally intensive that a billion divisions over a short time period is not outside the realm of possibility. Assuming that the 1 in 9 billion figure is correct and that results of different divisions are independent of one another, what is the probability that at least one error occurs in one billion divisions with this chip?

Individual A has a circle of five close friends (B, C, D, E, and F). A has heard a certain rumor from outside the circle and has invited the five friends to a party to circulate the rumor. To begin, A selects one of the five at random and tells the rumor to the chosen individual. That individual then selects at random one of the four remaining individuals and repeats the rumor. Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told. a. What is the probability that the rumor is repeated in the order \(\mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}\), and \(\mathrm{F}\) ? b. What is the probability that \(\mathrm{F}\) is the third person at the party to be told the rumor? c. What is the probability that \(\mathrm{F}\) is the last person to hear the rumor? d. If at each stage the person who currently "has" the rumor does not know who has already heard it and selects the next recipient at random from all five possible individuals, what is the probability that \(\mathrm{F}\) has still not heard the rumor after it has been told ten times at the party?

Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects. a. How many ways are there to randomly select 5 of these keyboards for a thorough inspection (without regard to order)? b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect? c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect?

Professor Stan der Deviation can take one of two routes on his way home from work. On the first route, there are four railroad crossings. The probability that he will be stopped by a train at any particular one of the crossings is .1, and trains operate independently at the four crossings. The other route is longer but there are only two crossings, independent of one another, with the same stoppage probability for each as on the first route. On a particular day, Professor Deviation has a meeting scheduled at home for a certain time. Whichever route he takes, he calculates that he will be late if he is stopped by trains at at least half the crossings encountered. a. Which route should he take to minimize the probability of being late to the meeting? b. If he tosses a fair coin to decide on a route and he is late, what is the probability that he took the four-crossing route?
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